Is it true that $a^{360m + 1} + b^{360n + 1} equiv 0 (mod 475) Leftrightarrow a + b equiv 0 (mod 475)$?












0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












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