Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?
$begingroup$
Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?
It is obvious that we only need to check if $a + b equiv 0 (mod 475)$ implied $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475)$ or not.
With the notice that $360 = varphi (475)$, we are done with the case $(a,475) = 1$ (which leads to $(b,475) = 1$).
But when dealing with the case $(a,475) = 5$, I have some doubt. In this case, $a = 5k$, $b = 5l$ where $(k,5) = 1$, $(l,5) = 1$. A counter-example where $a + b equiv 0 (mod 25)$ but $a^{360m + 1} + b^{360n + 1} notequiv 0 (mod 25)$ could have existed.
I'm not sure how to proceed as I don't even know if the statement itself was true at the beginning. Please give me some hint. Thank you.
elementary-number-theory
asked Dec 29 '18 at 3:45
ElementXElementX
404111
$endgroup$
|
show 3 more comments
$begingroup$
Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?
It is obvious that we only need to check if $a + b equiv 0 (mod 475)$ implied $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475)$ or not.
With the notice that $360 = varphi (475)$, we are done with the case $(a,475) = 1$ (which leads to $(b,475) = 1$).
But when dealing with the case $(a,475) = 5$, I have some doubt. In this case, $a = 5k$, $b = 5l$ where $(k,5) = 1$, $(l,5) = 1$. A counter-example where $a + b equiv 0 (mod 25)$ but $a^{360m + 1} + b^{360n + 1} notequiv 0 (mod 25)$ could have existed.
I'm not sure how to proceed as I don't even know if the statement itself was true at the beginning. Please give me some hint. Thank you.
elementary-number-theory
asked Dec 29 '18 at 3:45
ElementXElementX
404111
$endgroup$
$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
1
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
1
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
1
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19
|
show 3 more comments
$begingroup$
Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?
It is obvious that we only need to check if $a + b equiv 0 (mod 475)$ implied $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475)$ or not.
With the notice that $360 = varphi (475)$, we are done with the case $(a,475) = 1$ (which leads to $(b,475) = 1$).
But when dealing with the case $(a,475) = 5$, I have some doubt. In this case, $a = 5k$, $b = 5l$ where $(k,5) = 1$, $(l,5) = 1$. A counter-example where $a + b equiv 0 (mod 25)$ but $a^{360m + 1} + b^{360n + 1} notequiv 0 (mod 25)$ could have existed.
I'm not sure how to proceed as I don't even know if the statement itself was true at the beginning. Please give me some hint. Thank you.
elementary-number-theory
asked Dec 29 '18 at 3:45
ElementXElementX
404111
$endgroup$
Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?
It is obvious that we only need to check if $a + b equiv 0 (mod 475)$ implied $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475)$ or not.
With the notice that $360 = varphi (475)$, we are done with the case $(a,475) = 1$ (which leads to $(b,475) = 1$).
But when dealing with the case $(a,475) = 5$, I have some doubt. In this case, $a = 5k$, $b = 5l$ where $(k,5) = 1$, $(l,5) = 1$. A counter-example where $a + b equiv 0 (mod 25)$ but $a^{360m + 1} + b^{360n + 1} notequiv 0 (mod 25)$ could have existed.
I'm not sure how to proceed as I don't even know if the statement itself was true at the beginning. Please give me some hint. Thank you.
elementary-number-theory
elementary-number-theory
asked Dec 29 '18 at 3:45
ElementXElementX
404111
asked Dec 29 '18 at 3:45
ElementXElementX
404111
asked Dec 29 '18 at 3:45
ElementXElementX
404111
asked Dec 29 '18 at 3:45
ElementXElementX
404111
asked Dec 29 '18 at 3:45
ElementXElementX
404111
404111
$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
1
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
1
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
1
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19
|
show 3 more comments
$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
1
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
1
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
1
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19
$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
1
1
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
1
1
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
1
1
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19
|
show 3 more comments
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$begingroup$
Why is the first direction obvious? If you're using Fermat's little theorem then doesn't it just work the other way as well? I don't see the issue.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:22
1
$begingroup$
I use this: $a^{2k+1}+ b^{2k+1} = (a+b) .(...)$
$endgroup$
– ElementX
Dec 29 '18 at 4:28
$begingroup$
And for further comment, $475$ is not a prime for you to use Fermat's little theorem. I'm using Euler's theorem for the case $(a,475) = 1$. But when $(a,475) > 1$, Euler's theorem can't be used.
$endgroup$
– ElementX
Dec 29 '18 at 4:30
1
$begingroup$
I see. Well the general Fermat's little theorem says $a^{varphi(n)} equiv 1 pmod{n}$, and so $a^{varphi(n) + 1} equiv a pmod{n}$. Then the claim easily follows.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 4:31
1
$begingroup$
If $aequiv bpmod{n}$, then $$a^{360m+1}+b^{360m+1}equiv a^{360m+1}+(-a)^{360m+1}equiv a^{360m+1}+(-1)^{360m+1}a^{360m+1}equiv a^{360m+1}-a^{360m+1}equiv 0pmod{n}.$$ You seem to have the wrong idea about which implication is trivial, if I'm not mistaken.
$endgroup$
– saulspatz
Dec 29 '18 at 6:19