Challenge Level

Alan from Madras College, St Andrew's,
Scotland and Alexander from Shevah-Mofet School, Israel both used
induction to prove that $5^{2n+1} + 11^{2n+1} + 17^{2n+1}$ is
divisible by $33$ for all non-negative integer values of
*n* . This is
Alexander's proof.

Let us assume that $p =2n+1$, which means that *p* is a
positive odd number. In that case we need to prove that
$5^p+11^p+17^p$ divides by $33$.

Obviously if $n=0$, then $p=1$ and $5+11+17=33$ divides by $33$. Let's assume that $5^p +11^p +17^p$ divides by $33$ and prove that $5^{p+2} + 11^{p+2} + 17^{p+2}$ divides by $33$ which means we need to prove that $25 \times5^p+ 121 \times11^p + 289 \times17^p$ divides by $33$.

Since we know that $5^p +11^p +17^p$ divides by $33$, we can
subtract it $25$ times and we'll need to prove that $96 \times11^p+
264 \times17^p$ divides by $33$. Now $264 = 8 \times33$ which means
that $264 \times17^p$ divides by $33$. All that is left to prove is
that $96 \times11^p$ divides by $33$ and we're done. As $96$
divides by $3$ and $11^p$ divides by $11$ ( $p > 0$) so $96
\times11^p$ divides by $33$. We've proven that if $5^p +11^p +17^p$
divides by $33$ then $5^{p+2} + 11^{p+2} + 17^{p+2}$ also divides
by $33$ and we've checked that it is correct for $ p=1$. We have
proved inductively that $5^{2n+1} + 11^{2n+1} + 17^{2n+1}$ is
divisible by $33$ for all non-negative integer values of *$n$* .

This can also be proved using modulus arithmetic as follows:

$5^p + 11^p + 17^p \equiv (-6)^p + 0^p + 6^p \equiv 0$ (mod 11)

because $p$is odd, and $5^p + 11^p + 17^p \equiv 2^p + 2^p +
2^p \equiv 3 x 2^p \equiv 0$ (mod 3).

For more
discussion of solutions to this problem see the article.