Diophantine equation: solving $a^2+4n=b^2$
$begingroup$
I found myself working with diophantine equations but I have no experience at all with them.
Given an integer $n$, can I find two integers, $a$ and $b$, such that
$$a^2+4n=b^2$$
How would you guys approach the problem?
Thank you in advance.
diophantine-equations square-numbers sums-of-squares
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
$endgroup$
add a comment |
$begingroup$
I found myself working with diophantine equations but I have no experience at all with them.
Given an integer $n$, can I find two integers, $a$ and $b$, such that
$$a^2+4n=b^2$$
How would you guys approach the problem?
Thank you in advance.
diophantine-equations square-numbers sums-of-squares
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
$endgroup$
add a comment |
$begingroup$
I found myself working with diophantine equations but I have no experience at all with them.
Given an integer $n$, can I find two integers, $a$ and $b$, such that
$$a^2+4n=b^2$$
How would you guys approach the problem?
Thank you in advance.
diophantine-equations square-numbers sums-of-squares
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
$endgroup$
I found myself working with diophantine equations but I have no experience at all with them.
Given an integer $n$, can I find two integers, $a$ and $b$, such that
$$a^2+4n=b^2$$
How would you guys approach the problem?
Thank you in advance.
diophantine-equations square-numbers sums-of-squares
diophantine-equations square-numbers sums-of-squares
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
edited Nov 26 '18 at 12:50
José Carlos Santos
156k22125227
156k22125227
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
asked Nov 26 '18 at 12:34
Lyn CassidyLyn Cassidy
436
436
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Sure. With $a=n-1, b=n+1$ we have $$a^2+4n=n^2-2n+1+4n=n^2+2n+1=b^2$$
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
$endgroup$
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
add a comment |
$begingroup$
$$4n=b^2-a^2$$
$$n=dfrac{b+a}2cdotdfrac{b-a}2$$
If $n=pcdot q,dfrac{b+a}2=p, dfrac{b-a}2=q$
Trivially $q=1,p=?$
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
add a comment |
$begingroup$
Here is a method to compute $0leq a,bleq 1000$ given n.
public static long diophantine(long n)
{
long fourn= 4*n;
long results = new long[2][1000];
int index=0;
for(int a=1;a<1000;a++)
{
double b=Math.sqrt(Math.pow((double)a, 2.0)+(double)fourn);
if((b % 1) == 0)
{
results[0][index]=a;
results[1][index]=(long)b;
index++;
}
}
return results;
}
SO for example, when $n=50$ gives(first column is the values of a and the second is the values of b :
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
$endgroup$
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
add a comment |
$begingroup$
4 n = b^2 - a^2
n = g * h
4 n = (b - a) (b + a)
4 n = 2 g * 2 h
b - a = 2 g; b + a = 2 h
n = g h
a = h - g
b = h + g
if n = 7 then
(g=-7; h=-1 => a=6; b=-8 or
g=-1; h=-7 => a=-6; b=-8 or
g=1; h=7 => a=6; b=8 or
g=7; h=1 =>a=-6; b=8)
https://www.youtube.com/watch?v=dkf2xnmGHuA&index=8&list=PLfd6TTV3Dn5JUp6IDep5jV9wzOEmkpv4O
answered Nov 27 '18 at 15:38
S. I.S. I.
112
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sure. With $a=n-1, b=n+1$ we have $$a^2+4n=n^2-2n+1+4n=n^2+2n+1=b^2$$
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
$endgroup$
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
add a comment |
$begingroup$
Sure. With $a=n-1, b=n+1$ we have $$a^2+4n=n^2-2n+1+4n=n^2+2n+1=b^2$$
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
$endgroup$
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
add a comment |
$begingroup$
Sure. With $a=n-1, b=n+1$ we have $$a^2+4n=n^2-2n+1+4n=n^2+2n+1=b^2$$
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
$endgroup$
Sure. With $a=n-1, b=n+1$ we have $$a^2+4n=n^2-2n+1+4n=n^2+2n+1=b^2$$
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
answered Nov 26 '18 at 12:38
lulululu
39.8k24778
39.8k24778
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
add a comment |
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
Hi thank you too for your answer, are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:02
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
all the answers are trivial. Rewrite your equation as $4n=(b-a)(b+a)$. Write $4n=(2m)times (2k)$ for any factoring, $n=mk$ If $m≥k$ then solve $b+a=2m,b-a=2k$.
$endgroup$
– lulu
Nov 26 '18 at 13:07
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
$begingroup$
So basically to get all the results you need to factorize n. I'll mark this as the correct solution.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:27
add a comment |
$begingroup$
$$4n=b^2-a^2$$
$$n=dfrac{b+a}2cdotdfrac{b-a}2$$
If $n=pcdot q,dfrac{b+a}2=p, dfrac{b-a}2=q$
Trivially $q=1,p=?$
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
add a comment |
$begingroup$
$$4n=b^2-a^2$$
$$n=dfrac{b+a}2cdotdfrac{b-a}2$$
If $n=pcdot q,dfrac{b+a}2=p, dfrac{b-a}2=q$
Trivially $q=1,p=?$
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
add a comment |
$begingroup$
$$4n=b^2-a^2$$
$$n=dfrac{b+a}2cdotdfrac{b-a}2$$
If $n=pcdot q,dfrac{b+a}2=p, dfrac{b-a}2=q$
Trivially $q=1,p=?$
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$4n=b^2-a^2$$
$$n=dfrac{b+a}2cdotdfrac{b-a}2$$
If $n=pcdot q,dfrac{b+a}2=p, dfrac{b-a}2=q$
Trivially $q=1,p=?$
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 26 '18 at 12:40
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
add a comment |
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
$begingroup$
Hi, thanks for your answer. Are you aware of methods to find non-trivial solutions?
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:00
add a comment |
$begingroup$
Here is a method to compute $0leq a,bleq 1000$ given n.
public static long diophantine(long n)
{
long fourn= 4*n;
long results = new long[2][1000];
int index=0;
for(int a=1;a<1000;a++)
{
double b=Math.sqrt(Math.pow((double)a, 2.0)+(double)fourn);
if((b % 1) == 0)
{
results[0][index]=a;
results[1][index]=(long)b;
index++;
}
}
return results;
}
SO for example, when $n=50$ gives(first column is the values of a and the second is the values of b :
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
$endgroup$
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
add a comment |
$begingroup$
Here is a method to compute $0leq a,bleq 1000$ given n.
public static long diophantine(long n)
{
long fourn= 4*n;
long results = new long[2][1000];
int index=0;
for(int a=1;a<1000;a++)
{
double b=Math.sqrt(Math.pow((double)a, 2.0)+(double)fourn);
if((b % 1) == 0)
{
results[0][index]=a;
results[1][index]=(long)b;
index++;
}
}
return results;
}
SO for example, when $n=50$ gives(first column is the values of a and the second is the values of b :
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
$endgroup$
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
add a comment |
$begingroup$
Here is a method to compute $0leq a,bleq 1000$ given n.
public static long diophantine(long n)
{
long fourn= 4*n;
long results = new long[2][1000];
int index=0;
for(int a=1;a<1000;a++)
{
double b=Math.sqrt(Math.pow((double)a, 2.0)+(double)fourn);
if((b % 1) == 0)
{
results[0][index]=a;
results[1][index]=(long)b;
index++;
}
}
return results;
}
SO for example, when $n=50$ gives(first column is the values of a and the second is the values of b :
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
$endgroup$
Here is a method to compute $0leq a,bleq 1000$ given n.
public static long diophantine(long n)
{
long fourn= 4*n;
long results = new long[2][1000];
int index=0;
for(int a=1;a<1000;a++)
{
double b=Math.sqrt(Math.pow((double)a, 2.0)+(double)fourn);
if((b % 1) == 0)
{
results[0][index]=a;
results[1][index]=(long)b;
index++;
}
}
return results;
}
SO for example, when $n=50$ gives(first column is the values of a and the second is the values of b :
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
answered Nov 26 '18 at 13:05
nafhgoodnafhgood
1,801422
1,801422
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
add a comment |
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
$begingroup$
Hi, thanks for the effort, but this is what i was trying to avoid: no loops, just a straight up equation.
$endgroup$
– Lyn Cassidy
Nov 26 '18 at 13:24
add a comment |
$begingroup$
4 n = b^2 - a^2
n = g * h
4 n = (b - a) (b + a)
4 n = 2 g * 2 h
b - a = 2 g; b + a = 2 h
n = g h
a = h - g
b = h + g
if n = 7 then
(g=-7; h=-1 => a=6; b=-8 or
g=-1; h=-7 => a=-6; b=-8 or
g=1; h=7 => a=6; b=8 or
g=7; h=1 =>a=-6; b=8)
https://www.youtube.com/watch?v=dkf2xnmGHuA&index=8&list=PLfd6TTV3Dn5JUp6IDep5jV9wzOEmkpv4O
answered Nov 27 '18 at 15:38
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
4 n = b^2 - a^2
n = g * h
4 n = (b - a) (b + a)
4 n = 2 g * 2 h
b - a = 2 g; b + a = 2 h
n = g h
a = h - g
b = h + g
if n = 7 then
(g=-7; h=-1 => a=6; b=-8 or
g=-1; h=-7 => a=-6; b=-8 or
g=1; h=7 => a=6; b=8 or
g=7; h=1 =>a=-6; b=8)
https://www.youtube.com/watch?v=dkf2xnmGHuA&index=8&list=PLfd6TTV3Dn5JUp6IDep5jV9wzOEmkpv4O
answered Nov 27 '18 at 15:38
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
4 n = b^2 - a^2
n = g * h
4 n = (b - a) (b + a)
4 n = 2 g * 2 h
b - a = 2 g; b + a = 2 h
n = g h
a = h - g
b = h + g
if n = 7 then
(g=-7; h=-1 => a=6; b=-8 or
g=-1; h=-7 => a=-6; b=-8 or
g=1; h=7 => a=6; b=8 or
g=7; h=1 =>a=-6; b=8)
https://www.youtube.com/watch?v=dkf2xnmGHuA&index=8&list=PLfd6TTV3Dn5JUp6IDep5jV9wzOEmkpv4O
answered Nov 27 '18 at 15:38
S. I.S. I.
112
$endgroup$
4 n = b^2 - a^2
n = g * h
4 n = (b - a) (b + a)
4 n = 2 g * 2 h
b - a = 2 g; b + a = 2 h
n = g h
a = h - g
b = h + g
if n = 7 then
(g=-7; h=-1 => a=6; b=-8 or
g=-1; h=-7 => a=-6; b=-8 or
g=1; h=7 => a=6; b=8 or
g=7; h=1 =>a=-6; b=8)
https://www.youtube.com/watch?v=dkf2xnmGHuA&index=8&list=PLfd6TTV3Dn5JUp6IDep5jV9wzOEmkpv4O
answered Nov 27 '18 at 15:38
S. I.S. I.
112
edited Nov 27 '18 at 15:50
answered Nov 27 '18 at 15:38
S. I.S. I.
112
answered Nov 27 '18 at 15:38
S. I.S. I.
112
answered Nov 27 '18 at 15:38
S. I.S. I.
112
112
add a comment |
add a comment |
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