We can see that if $p=2$ then $a^n = 13$ $\implies$ $n=1$. Now, we can use LTE with odd exponents (as $p$ is odd unless $p = 2$). Then $v_5(2^p+3^p) = v_5(2+3) + v_5(p) = 1$ unless $p=5$ in which case $a^n=275$ which is only a perfect first power. Otherwise $v_5(a^n) = 1$ $\implies$ $n=1$ I am guessing that LTE was not very well known at the time because this is obviously too easy for a TST P6; but the problem is probably much more difficult without LTE (even though one can easily conclude that $5 \mid a$ after factorizing LHS).