Integral solutions of polynomial [duplicate]
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This question already has an answer here:
integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots.
2 answers
The equation
$$
a_n x^n+dots+a_0=1, qquad a_nneq 0, qquad a_iinmathbb{Z}
$$
has $geq 4$ distinct integral solutions. How can I prove that
$$
a_n x^n+dots+a_0=-1
$$
has no integral solution?
elementary-number-theory polynomials
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marked as duplicate by José Carlos Santos, Bill Dubuque elementary-number-theory
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Nov 23 '18 at 19:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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This question already has an answer here:
integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots.
2 answers
The equation
$$
a_n x^n+dots+a_0=1, qquad a_nneq 0, qquad a_iinmathbb{Z}
$$
has $geq 4$ distinct integral solutions. How can I prove that
$$
a_n x^n+dots+a_0=-1
$$
has no integral solution?
elementary-number-theory polynomials
$endgroup$
marked as duplicate by José Carlos Santos, Bill Dubuque elementary-number-theory
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Nov 23 '18 at 19:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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It does have an integral solution. Take x = 0 and a_0 = -1.
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– D.B.
Nov 23 '18 at 18:47
2
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@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
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– user285001
Nov 23 '18 at 18:48
add a comment |
$begingroup$
This question already has an answer here:
integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots.
2 answers
The equation
$$
a_n x^n+dots+a_0=1, qquad a_nneq 0, qquad a_iinmathbb{Z}
$$
has $geq 4$ distinct integral solutions. How can I prove that
$$
a_n x^n+dots+a_0=-1
$$
has no integral solution?
elementary-number-theory polynomials
$endgroup$
This question already has an answer here:
integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots.
2 answers
The equation
$$
a_n x^n+dots+a_0=1, qquad a_nneq 0, qquad a_iinmathbb{Z}
$$
has $geq 4$ distinct integral solutions. How can I prove that
$$
a_n x^n+dots+a_0=-1
$$
has no integral solution?
This question already has an answer here:
integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots.
2 answers
elementary-number-theory polynomials
elementary-number-theory polynomials
edited Nov 23 '18 at 18:49
asked Nov 23 '18 at 18:42
user285001
marked as duplicate by José Carlos Santos, Bill Dubuque elementary-number-theory
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Nov 23 '18 at 19:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Bill Dubuque elementary-number-theory
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Nov 23 '18 at 19:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
It does have an integral solution. Take x = 0 and a_0 = -1.
$endgroup$
– D.B.
Nov 23 '18 at 18:47
2
$begingroup$
@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
$endgroup$
– user285001
Nov 23 '18 at 18:48
add a comment |
$begingroup$
It does have an integral solution. Take x = 0 and a_0 = -1.
$endgroup$
– D.B.
Nov 23 '18 at 18:47
2
$begingroup$
@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
$endgroup$
– user285001
Nov 23 '18 at 18:48
$begingroup$
It does have an integral solution. Take x = 0 and a_0 = -1.
$endgroup$
– D.B.
Nov 23 '18 at 18:47
$begingroup$
It does have an integral solution. Take x = 0 and a_0 = -1.
$endgroup$
– D.B.
Nov 23 '18 at 18:47
2
2
$begingroup$
@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
$endgroup$
– user285001
Nov 23 '18 at 18:48
$begingroup$
@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
$endgroup$
– user285001
Nov 23 '18 at 18:48
add a comment |
1 Answer
1
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oldest
votes
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Hint:
Let $f(x)=sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$
Then,
$$f(x)=(x-p)(x-q)(x-r)(x-s) ,,g(x)+1, quad text{where } g(x) text{ is some polynomial in } Bbb{Z}[x].$$
If $t in Bbb{Z}$ is such that $f(t)=-1$. Then,
$$f(t)=-1=(t-p)(t-q)(t-r)(t-s) ,,g(t)+1.$$
This is same as saying
$$(t-p)(t-q)(t-r)(t-s) ,,g(t)=-2$$
Observe that $t notin{p,q,r,s}$ and the left side is a product of integers. Can you now arrive at a contradiction?
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Let $f(x)=sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$
Then,
$$f(x)=(x-p)(x-q)(x-r)(x-s) ,,g(x)+1, quad text{where } g(x) text{ is some polynomial in } Bbb{Z}[x].$$
If $t in Bbb{Z}$ is such that $f(t)=-1$. Then,
$$f(t)=-1=(t-p)(t-q)(t-r)(t-s) ,,g(t)+1.$$
This is same as saying
$$(t-p)(t-q)(t-r)(t-s) ,,g(t)=-2$$
Observe that $t notin{p,q,r,s}$ and the left side is a product of integers. Can you now arrive at a contradiction?
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Hint:
Let $f(x)=sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$
Then,
$$f(x)=(x-p)(x-q)(x-r)(x-s) ,,g(x)+1, quad text{where } g(x) text{ is some polynomial in } Bbb{Z}[x].$$
If $t in Bbb{Z}$ is such that $f(t)=-1$. Then,
$$f(t)=-1=(t-p)(t-q)(t-r)(t-s) ,,g(t)+1.$$
This is same as saying
$$(t-p)(t-q)(t-r)(t-s) ,,g(t)=-2$$
Observe that $t notin{p,q,r,s}$ and the left side is a product of integers. Can you now arrive at a contradiction?
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Hint:
Let $f(x)=sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$
Then,
$$f(x)=(x-p)(x-q)(x-r)(x-s) ,,g(x)+1, quad text{where } g(x) text{ is some polynomial in } Bbb{Z}[x].$$
If $t in Bbb{Z}$ is such that $f(t)=-1$. Then,
$$f(t)=-1=(t-p)(t-q)(t-r)(t-s) ,,g(t)+1.$$
This is same as saying
$$(t-p)(t-q)(t-r)(t-s) ,,g(t)=-2$$
Observe that $t notin{p,q,r,s}$ and the left side is a product of integers. Can you now arrive at a contradiction?
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
$endgroup$
Hint:
Let $f(x)=sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$
Then,
$$f(x)=(x-p)(x-q)(x-r)(x-s) ,,g(x)+1, quad text{where } g(x) text{ is some polynomial in } Bbb{Z}[x].$$
If $t in Bbb{Z}$ is such that $f(t)=-1$. Then,
$$f(t)=-1=(t-p)(t-q)(t-r)(t-s) ,,g(t)+1.$$
This is same as saying
$$(t-p)(t-q)(t-r)(t-s) ,,g(t)=-2$$
Observe that $t notin{p,q,r,s}$ and the left side is a product of integers. Can you now arrive at a contradiction?
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
answered Nov 23 '18 at 19:11
Anurag AAnurag A
25.8k12249
25.8k12249
add a comment |
add a comment |
$begingroup$
It does have an integral solution. Take x = 0 and a_0 = -1.
$endgroup$
– D.B.
Nov 23 '18 at 18:47
2
$begingroup$
@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$
$endgroup$
– user285001
Nov 23 '18 at 18:48