If $8x^p+1$ is a square then odd prime p divides x-1
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For odd prime $p$ and positive integer $x$, prove that if $8x^p+1$ is a perfect square then odd prime $p$ divides $x-1$.
I've tried modulo $p$ and factorizing with $8x^p=k^2-1$. It gives some sort of equation like $a^p-2b^p=1$ which does not help. I also tried $x=pk+1+r$ but didn't go well. The most curious part of this problem is that there seems to be no relationship between $x-1$ and $p$ (perhaps $x$ or $p$ maybe).
number-theory elementary-number-theory
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
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show 2 more comments
$begingroup$
For odd prime $p$ and positive integer $x$, prove that if $8x^p+1$ is a perfect square then odd prime $p$ divides $x-1$.
I've tried modulo $p$ and factorizing with $8x^p=k^2-1$. It gives some sort of equation like $a^p-2b^p=1$ which does not help. I also tried $x=pk+1+r$ but didn't go well. The most curious part of this problem is that there seems to be no relationship between $x-1$ and $p$ (perhaps $x$ or $p$ maybe).
number-theory elementary-number-theory
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
$endgroup$
1
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
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Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
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This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
1
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
1
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24
|
show 2 more comments
$begingroup$
For odd prime $p$ and positive integer $x$, prove that if $8x^p+1$ is a perfect square then odd prime $p$ divides $x-1$.
I've tried modulo $p$ and factorizing with $8x^p=k^2-1$. It gives some sort of equation like $a^p-2b^p=1$ which does not help. I also tried $x=pk+1+r$ but didn't go well. The most curious part of this problem is that there seems to be no relationship between $x-1$ and $p$ (perhaps $x$ or $p$ maybe).
number-theory elementary-number-theory
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
$endgroup$
For odd prime $p$ and positive integer $x$, prove that if $8x^p+1$ is a perfect square then odd prime $p$ divides $x-1$.
I've tried modulo $p$ and factorizing with $8x^p=k^2-1$. It gives some sort of equation like $a^p-2b^p=1$ which does not help. I also tried $x=pk+1+r$ but didn't go well. The most curious part of this problem is that there seems to be no relationship between $x-1$ and $p$ (perhaps $x$ or $p$ maybe).
number-theory elementary-number-theory
number-theory elementary-number-theory
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
edited Dec 30 '18 at 19:09
Hypernova
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
asked Dec 30 '18 at 15:56
HypernovaHypernova
83
83
1
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
$begingroup$
Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
$begingroup$
This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
1
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
1
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24
|
show 2 more comments
1
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
$begingroup$
Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
$begingroup$
This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
1
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
1
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24
1
1
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
$begingroup$
Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
$begingroup$
Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
$begingroup$
This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
$begingroup$
This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
1
1
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
1
1
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24
|
show 2 more comments
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1
$begingroup$
What have you tried?
$endgroup$
– Lucas Henrique
Dec 30 '18 at 15:58
$begingroup$
Please include more context via an edit. More details are in the banner and the linked thread there.
$endgroup$
– quid♦
Dec 30 '18 at 16:00
$begingroup$
This is false btw. Let $p=3,x=83232$ then $sqrt{8x^3+1} = 67917312$ and $p|x$ thus $pnmid x-1$
$endgroup$
– Anvit
Dec 30 '18 at 16:25
1
$begingroup$
@above But the square root of $8x^3+1$ should always be a odd number. Something's wrong with your counterexample.
$endgroup$
– Hypernova
Dec 30 '18 at 19:11
1
$begingroup$
Do you have any examples of perfect squares of the form $8x^p+1$ with $p$ an odd prime, never mind whether $p | x-1$? Notice that if $8n+1$ is a perfect square $s^2$ then $n$ is a triangle number since $n = [(s/2-1/2)(s/2+1/2)/2]$. But answers to this question strongly suggest a triangle number cannot be a perfect power with exponent greater than two.
$endgroup$
– Adam Bailey
Dec 30 '18 at 21:24