Find all triples $(p,x,y)$ such that $ p^x=y^4+4$
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Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 equiv (y-1)^2 equiv -1 pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
He then says that by$mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
- If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k equiv p^j equiv 2 pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
- Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
- $ 2y^{2}+4=p^{b}+p^{a}$, if $ aneq 0$, then $ y^2 equiv 3mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
- $gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
number-theory diophantine-equations quadratics
$endgroup$
add a comment |
$begingroup$
Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 equiv (y-1)^2 equiv -1 pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
He then says that by$mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
- If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k equiv p^j equiv 2 pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
- Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
- $ 2y^{2}+4=p^{b}+p^{a}$, if $ aneq 0$, then $ y^2 equiv 3mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
- $gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
number-theory diophantine-equations quadratics
$endgroup$
$begingroup$
Can you put a link to the original answer you want to understand?
$endgroup$
– xarles
Aug 30 '18 at 17:05
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For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
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@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
$endgroup$
– Vmimi
Aug 31 '18 at 19:31
$begingroup$
In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
$endgroup$
– Vmimi
Sep 3 '18 at 23:05
add a comment |
$begingroup$
Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 equiv (y-1)^2 equiv -1 pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
He then says that by$mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
- If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k equiv p^j equiv 2 pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
- Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
- $ 2y^{2}+4=p^{b}+p^{a}$, if $ aneq 0$, then $ y^2 equiv 3mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
- $gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
number-theory diophantine-equations quadratics
$endgroup$
Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x = (y^2+2y+2)(y^2-2y+2)$
Then: $p^k = y^2+2y+2$ and $p^j= y^2-2y+2$
The solutions I saw were different from here and they just wanted to prove that $(5,1,1)$ was the only solution. I´ll just ask what I didn't understand:
1)
Therefore, $p^k = (y+1)^2+1$ and $p^j= (y-1)^2+1$
And then: $(y+1)^2 equiv (y-1)^2 equiv -1 pmod p$
So here $-1$ is a quadratic residue, and therefore $p= 4n+1$. I understood his solution until this part.
I know that $p= 4n+1$ because I saw that was one of the properties of quadratic residues, but would like to see a proof of it if possible.
He then says that by$mod 8$ , $ k$ and $j$ had different parity, so $x$ was odd.
Then he assumes that $p, k >1$ and says:
- If $k=2m$, then: $(y+1)^2= (p^m+1)(p^m -1)$ or if $j= 2m$: then $(y-1)^2= (p^m+1)(p^m -1)$ and affirms those equations have no solution, that might be a property but want to know about it anyway.
Then he says there are no solution for $x>1$, then $p = (y^2+2y+2)(y^2-2y+2)$, so $1=y^2-2y+2$ and that was the triple $(5,1,1)$, therefore that is the only solution.
2)
He started off by saying: if $y$ is even then $ p^k equiv p^j equiv 2 pmod 4$ then $a = b =1$ and that is not possible, therefore $y$ is odd.
- Since $ 4y = p^{b} - p^{a}$, therefore $ p^{b-a}=5$, $ p^{a}=y$
- $ 2y^{2}+4=p^{b}+p^{a}$, if $ aneq 0$, then $ y^2 equiv 3mod 5$, which is impossible.
I don´t know how you can deduce those things. After that he affirmed that $(5,1,1)$ was the only solution.
3) He showed that $y$ was odd and said:
- $gcd (y^2+2y+2, y^2-2y+2) = gcd (y^2+2y+2, 4y) = 1$ (I think this is also a property but I don´t even know it).
And since $y^2+2y+2 > y^2-2y+2$, then $y^2+2y+2=p^x$ and $y^2-2y+2 =1$, so $y=1$ and $(5,1,1)$ was the only solution.
Sorry if I asked many things, I just thought those things may be useful for other problems. I pointed out the things I didn´t understood so that it was easy to see them. Thanks in advance.
number-theory diophantine-equations quadratics
number-theory diophantine-equations quadratics
asked Aug 30 '18 at 2:42
VmimiVmimi
372212
372212
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Can you put a link to the original answer you want to understand?
$endgroup$
– xarles
Aug 30 '18 at 17:05
$begingroup$
For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
$begingroup$
@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
$endgroup$
– Vmimi
Aug 31 '18 at 19:31
$begingroup$
In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
$endgroup$
– Vmimi
Sep 3 '18 at 23:05
add a comment |
$begingroup$
Can you put a link to the original answer you want to understand?
$endgroup$
– xarles
Aug 30 '18 at 17:05
$begingroup$
For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
$begingroup$
@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
$endgroup$
– Vmimi
Aug 31 '18 at 19:31
$begingroup$
In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
$endgroup$
– Vmimi
Sep 3 '18 at 23:05
$begingroup$
Can you put a link to the original answer you want to understand?
$endgroup$
– xarles
Aug 30 '18 at 17:05
$begingroup$
Can you put a link to the original answer you want to understand?
$endgroup$
– xarles
Aug 30 '18 at 17:05
$begingroup$
For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
$begingroup$
For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
$begingroup$
@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
$endgroup$
– Vmimi
Aug 31 '18 at 19:31
$begingroup$
@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
$endgroup$
– Vmimi
Aug 31 '18 at 19:31
$begingroup$
In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
$endgroup$
– Vmimi
Sep 3 '18 at 23:05
$begingroup$
In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
$endgroup$
– Vmimi
Sep 3 '18 at 23:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y geq 6$.
So we conclude e.g., $y le 5$ and $p^x le 5^4+4$. Furthermore using the above reasoning you can conclude $p le 5$ if $y in {2,3,4,5}$. This leaves us only a very small set of triples to check via brute force.
$endgroup$
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
add a comment |
$begingroup$
Alternative Solution:
First, assume that $'y'$ is even , this case can be rejected easily.
Hence, $y$ is odd.
Note that if:
$p^{a}=(k^{2}+1)$ $,$ $k∈ℕ$
As $(k^{2}+1)$ can't be a perfect square , it clearly implies that $'a'$ is odd.(unless $k=0$ )
Now, $p^{i}=(y-1)^{2}+1$ ; $p^{j}=(y+1)^{2}+1$
(if $(y-1)=0$, we get the case $(5,1,1)$ )
(if $(y-1),(y+1)∈ℕ$ , then both $i,j$ are odd )
$p^{x}=p^{i}p^{j}=p^{i+j}$
(As $i,j$ are both odd , $(i+j)=x$ is even )
Putting $x=2k$ in original equation,
$p^{2k}=y^{4}+4$
$(p^{k}-y^{2})(p^{k}+y^{2})=4$
using the fact that $(p^{k}-y^{2})$ and $(p^{k}+y^{2})$ are natural number factors of $4$, and,
$(p^{k}+y^{2})>(p^{k}-y^{2})$
$(p^{k}+y^{2})=4$ ; $(p^{k}-y^{2})=1$
we can see that this is not possible as $ y^{2}≠1.5 $
Hence no possible solutions when $y≠1$
The only possible solution:
$(p,x,y)=(5,1,1)$
$endgroup$
$begingroup$
Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y geq 6$.
So we conclude e.g., $y le 5$ and $p^x le 5^4+4$. Furthermore using the above reasoning you can conclude $p le 5$ if $y in {2,3,4,5}$. This leaves us only a very small set of triples to check via brute force.
$endgroup$
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
add a comment |
$begingroup$
There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y geq 6$.
So we conclude e.g., $y le 5$ and $p^x le 5^4+4$. Furthermore using the above reasoning you can conclude $p le 5$ if $y in {2,3,4,5}$. This leaves us only a very small set of triples to check via brute force.
$endgroup$
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
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– Mike
Sep 1 '18 at 17:05
add a comment |
$begingroup$
There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y geq 6$.
So we conclude e.g., $y le 5$ and $p^x le 5^4+4$. Furthermore using the above reasoning you can conclude $p le 5$ if $y in {2,3,4,5}$. This leaves us only a very small set of triples to check via brute force.
$endgroup$
There is a MUCH easier way to finish.
Starting from $p^k=y^2+2y+2$; $p^j=y^2-2y+2$ we note on the one hand, the following inequality $y^2+2y+2 > y^2-2y+2$ (for natural numbers $y$) and so $j<k$, and therefore
$y^2+2y+2$ must be of the form $p^i(y^2-2y+2)$ for some prime and some positive integer $i$.
On the other hand, we note the following: $y^2+2y+2 < 2(y^2-2y+2)$ for all $y geq 6$.
So we conclude e.g., $y le 5$ and $p^x le 5^4+4$. Furthermore using the above reasoning you can conclude $p le 5$ if $y in {2,3,4,5}$. This leaves us only a very small set of triples to check via brute force.
edited Dec 6 '18 at 21:17
answered Aug 30 '18 at 19:39
MikeMike
4,206412
4,206412
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How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
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– Vmimi
Aug 31 '18 at 19:47
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If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
add a comment |
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
How can you show that $ p le 5$? by your reasoning $p^x < 5^4+1$ but I think you can´t deduce anything from there if you don´t have any information about $x$.
$endgroup$
– Vmimi
Aug 31 '18 at 19:47
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then $y^2+2y+2 = p^i(y^2-2y+2) le 5(y^2-2y+2)$; $i$ a positive integer , which implies $p$ must be no greater than 5.
$endgroup$
– Mike
Sep 1 '18 at 16:30
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
$begingroup$
If $y$ is at least 2 then on the one hand $y^2+2y+2 le 5(y^2-2y+2)$. However, on the other hand, it has already been shown that if $(p,x,y)$ is a triplet, that $y^2+2y+2 = p^i(y^2-2y+2)$ for some positive integer $i$. Thus we conclude that $p^i le 5$ for that positive integer $i$, which of course implies $p le 5$.
$endgroup$
– Mike
Sep 1 '18 at 17:05
add a comment |
$begingroup$
Alternative Solution:
First, assume that $'y'$ is even , this case can be rejected easily.
Hence, $y$ is odd.
Note that if:
$p^{a}=(k^{2}+1)$ $,$ $k∈ℕ$
As $(k^{2}+1)$ can't be a perfect square , it clearly implies that $'a'$ is odd.(unless $k=0$ )
Now, $p^{i}=(y-1)^{2}+1$ ; $p^{j}=(y+1)^{2}+1$
(if $(y-1)=0$, we get the case $(5,1,1)$ )
(if $(y-1),(y+1)∈ℕ$ , then both $i,j$ are odd )
$p^{x}=p^{i}p^{j}=p^{i+j}$
(As $i,j$ are both odd , $(i+j)=x$ is even )
Putting $x=2k$ in original equation,
$p^{2k}=y^{4}+4$
$(p^{k}-y^{2})(p^{k}+y^{2})=4$
using the fact that $(p^{k}-y^{2})$ and $(p^{k}+y^{2})$ are natural number factors of $4$, and,
$(p^{k}+y^{2})>(p^{k}-y^{2})$
$(p^{k}+y^{2})=4$ ; $(p^{k}-y^{2})=1$
we can see that this is not possible as $ y^{2}≠1.5 $
Hence no possible solutions when $y≠1$
The only possible solution:
$(p,x,y)=(5,1,1)$
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Why do you say $i,j$ are both odd?
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– Vmimi
Sep 3 '18 at 2:33
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@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
add a comment |
$begingroup$
Alternative Solution:
First, assume that $'y'$ is even , this case can be rejected easily.
Hence, $y$ is odd.
Note that if:
$p^{a}=(k^{2}+1)$ $,$ $k∈ℕ$
As $(k^{2}+1)$ can't be a perfect square , it clearly implies that $'a'$ is odd.(unless $k=0$ )
Now, $p^{i}=(y-1)^{2}+1$ ; $p^{j}=(y+1)^{2}+1$
(if $(y-1)=0$, we get the case $(5,1,1)$ )
(if $(y-1),(y+1)∈ℕ$ , then both $i,j$ are odd )
$p^{x}=p^{i}p^{j}=p^{i+j}$
(As $i,j$ are both odd , $(i+j)=x$ is even )
Putting $x=2k$ in original equation,
$p^{2k}=y^{4}+4$
$(p^{k}-y^{2})(p^{k}+y^{2})=4$
using the fact that $(p^{k}-y^{2})$ and $(p^{k}+y^{2})$ are natural number factors of $4$, and,
$(p^{k}+y^{2})>(p^{k}-y^{2})$
$(p^{k}+y^{2})=4$ ; $(p^{k}-y^{2})=1$
we can see that this is not possible as $ y^{2}≠1.5 $
Hence no possible solutions when $y≠1$
The only possible solution:
$(p,x,y)=(5,1,1)$
$endgroup$
$begingroup$
Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
add a comment |
$begingroup$
Alternative Solution:
First, assume that $'y'$ is even , this case can be rejected easily.
Hence, $y$ is odd.
Note that if:
$p^{a}=(k^{2}+1)$ $,$ $k∈ℕ$
As $(k^{2}+1)$ can't be a perfect square , it clearly implies that $'a'$ is odd.(unless $k=0$ )
Now, $p^{i}=(y-1)^{2}+1$ ; $p^{j}=(y+1)^{2}+1$
(if $(y-1)=0$, we get the case $(5,1,1)$ )
(if $(y-1),(y+1)∈ℕ$ , then both $i,j$ are odd )
$p^{x}=p^{i}p^{j}=p^{i+j}$
(As $i,j$ are both odd , $(i+j)=x$ is even )
Putting $x=2k$ in original equation,
$p^{2k}=y^{4}+4$
$(p^{k}-y^{2})(p^{k}+y^{2})=4$
using the fact that $(p^{k}-y^{2})$ and $(p^{k}+y^{2})$ are natural number factors of $4$, and,
$(p^{k}+y^{2})>(p^{k}-y^{2})$
$(p^{k}+y^{2})=4$ ; $(p^{k}-y^{2})=1$
we can see that this is not possible as $ y^{2}≠1.5 $
Hence no possible solutions when $y≠1$
The only possible solution:
$(p,x,y)=(5,1,1)$
$endgroup$
Alternative Solution:
First, assume that $'y'$ is even , this case can be rejected easily.
Hence, $y$ is odd.
Note that if:
$p^{a}=(k^{2}+1)$ $,$ $k∈ℕ$
As $(k^{2}+1)$ can't be a perfect square , it clearly implies that $'a'$ is odd.(unless $k=0$ )
Now, $p^{i}=(y-1)^{2}+1$ ; $p^{j}=(y+1)^{2}+1$
(if $(y-1)=0$, we get the case $(5,1,1)$ )
(if $(y-1),(y+1)∈ℕ$ , then both $i,j$ are odd )
$p^{x}=p^{i}p^{j}=p^{i+j}$
(As $i,j$ are both odd , $(i+j)=x$ is even )
Putting $x=2k$ in original equation,
$p^{2k}=y^{4}+4$
$(p^{k}-y^{2})(p^{k}+y^{2})=4$
using the fact that $(p^{k}-y^{2})$ and $(p^{k}+y^{2})$ are natural number factors of $4$, and,
$(p^{k}+y^{2})>(p^{k}-y^{2})$
$(p^{k}+y^{2})=4$ ; $(p^{k}-y^{2})=1$
we can see that this is not possible as $ y^{2}≠1.5 $
Hence no possible solutions when $y≠1$
The only possible solution:
$(p,x,y)=(5,1,1)$
edited Aug 30 '18 at 19:35
answered Aug 30 '18 at 19:15
kadoodlekadoodle
1297
1297
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Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
add a comment |
$begingroup$
Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
$begingroup$
Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
Why do you say $i,j$ are both odd?
$endgroup$
– Vmimi
Sep 3 '18 at 2:33
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
$begingroup$
@Vmimi After we say that y is odd, see the first thing that is written. If $p^{a}=k^{2}+1$, then, as $(k^{2}+1)$ can't be a perfect square (unless $k$ is $0$), hence either $k=0$ or else, $a=odd$. In the same way I say that either $i,j$ both are odd, or else, $(y-1)=0$.
$endgroup$
– kadoodle
Sep 3 '18 at 5:17
add a comment |
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Can you put a link to the original answer you want to understand?
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– xarles
Aug 30 '18 at 17:05
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For 3), use that $y^2+y+2-(y^2-2y+2)=4y$, and a prime that divides both, then divides $4y$, so it is $2$, which cannot be since $y$ is odd, or it divides $y$ and it divides $y(y+2)+2$, so $2$, which it was previously descarted.
$endgroup$
– xarles
Aug 30 '18 at 17:15
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@SofieVerbeek Thanks for your comment, I helped me understand some things, it´s a pity you deleted it. :(
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– Vmimi
Aug 31 '18 at 19:31
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In 1) he says that those equations don't have any solution because the gap between two perfect squares can't be one?
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– Vmimi
Sep 3 '18 at 23:05