A question from an old french olympiad

Asked Apr 19 '18 at 22:35

Active Apr 19 '18 at 23:03

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I've found an exercice that I've no clue to do it, I expose it to you

Let suppose we have a family of $1789$ integers which sum is equal to zero. Show that the sum of their $37$th power is a multiple of $399$.

Any hint or solution ?

edited Apr 19 '18 at 23:03

Thomas Andrews

asked Apr 19 '18 at 22:35

Atmos

2

Hint: look up the Carmichael Lambda Function and compute $\lambda (399)$ – lulu Apr 19 '18 at 22:44
@lulu $\lambda (399) = 18$, as calculated in https://comnuan.com/cmnn02/cmnn02006/cmnn02006.php
Now, how does it help to solve the problem?
– SARTHAK GUPTA Jul 28 '19 at 00:00
@SARTHAKGUPTA Well, since $\lambda(399)=18$ we have that $a^{36}\equiv \pmod {399}=1$ if $\gcd(399,a)=1$. Thus for those $a$ we have $a^{37}\equiv a \pmod {399}$. Now just check $2,3,19$ separately. – lulu Jul 28 '19 at 01:46

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