1

I was hoping someone could review my proof, thanks!

Question: For any integer $a$, $a$$^{37}$ $\equiv$ $a$ (mod 1729)

Proof:

Note first that 1729 = 7 * 13 * 19.

Then we have $a$$^{37}$ $\equiv$ $a$ mod 1729 if and only if the following set of equations all hold true for $a$:

$a$$^{37}$ $\equiv$ $a$ mod 7

$a$$^{37}$ $\equiv$ $a$ mod 13

$a$$^{37}$ $\equiv$ $a$ mod 19

For a proof of the reverse direction of the above if and only if statement, consider that if $a$ satisfies all 3 equations, then since all 3 numbers are relatively prime we have $a$$^{37}$ $\equiv$ $a$ mod lcm(7,13,19) = 1729.

Hence note that by Eulers Theorem we have a$^{\phi(7)}$ = a$^6$ $\equiv$ 1 mod 7, which implies a$^{37}$ $\equiv$ $a$ mod 7, for all $a$ such that gcd($a$,7) = 1. Similarly we have $a$$^{37}$ $\equiv$ $a$ mod 13 and $a$$^{37}$ $\equiv$ $a$ mod 19. Hence when $a$ satisfies the condition gcd($a$,7) = gcd($a$,13) = gcd($a$,19) = 1, then $a$$^{37}$ holds.

Suppose $a$ is divisible by either 7, 13, or 19. Then WLOG suppose $a$ $\equiv$ 0 mod 7. a$^{37}$ $\equiv$ $a$ mod 1729 if and only if the above 3 equations hold. The first equation holds as 0$^{37}$ $\equiv$ 0 mod 7. The other 2 equations also hold as gcd(7,13) = 1 and gcd(7,19) = 1, so Euler's Theorem still applies in those cases. Hence the original claim holds for all a $\in$ $\mathbb{Z}$.

Bill Dubuque
  • 282,220
H_1317
  • 1,125
  • 1
    Are you familiar with the Carmichael function ? – Peter Mar 07 '20 at 20:24
  • 2
    Of course, the approach using the chinese remainder theorem is good as well. – Peter Mar 07 '20 at 20:25
  • "Then we have $a^{37}$ is prime if and only if..." How can $a^{37}$ be prime? – fleablood Mar 07 '20 at 20:48
  • @fleablood , that was a typo. Thanks – H_1317 Mar 07 '20 at 20:53
  • Not sure I follow your reasoning. Why not just point out but FLT then $a^{kp+1}\equiv a\pmod p$ for any prime $p$ and integer $k$ so $a^{37}\equiv a\pmod m$ for $m=7,13,19$. So by chinese remainder theorem $a^{37}\equiv a\pmod{71319}$. – fleablood Mar 07 '20 at 20:54
  • Which part of my proof is not clear to you @fleablood – H_1317 Mar 07 '20 at 21:08
  • Some remarks: the "reverse direction" is CCRT = Constant-case CRT, It is not true that $,1729\mid a,$ if $a$ is divisible by $7,13$ or $19$. See the linked dupe for how to fix that. – Bill Dubuque Mar 07 '20 at 21:34
  • @BillDubuque, I tried to fix the last part myself, does that work? – H_1317 Mar 07 '20 at 22:12
  • @H_1317 It's still wrong because $a$ might be divisible by more than one prime. The easiest way to handle these cases is the way I do it in the linked dupe. – Bill Dubuque Mar 07 '20 at 23:05
  • Though the same argument still “essentially” holds: if divisible by any combination of elements of the set {7,13,19}, then we get 0 mod n for the equations where a is divisible by some element of the set and the gcd argument works for any remaining factor @BillDubuque – H_1317 Mar 07 '20 at 23:09
  • Yes, but it's clearer to do it all locally for each prime than to scatter it about. It is important to learn to think that way since reducing to prime(power) cases is uniquitous in number theory (CRT, local - global techniques, p-adics, etc) – Bill Dubuque Mar 07 '20 at 23:19

0 Answers0