First round – City competition

3.2. If we write down the given equation in the form $2^m{p}^2=n^5-1$ and factorise the right-hand side, we get $$2^m{p}^2=(n-1)(n^4+n^3+n^2+n+1).$$ Factor $n^4+n^3+n^2+n+1$ is odd, so $n-1$ is divisible by $2^m$. We immediately see that $p$ is odd. On the other hand, since $n$ is positive, we clearly have $n^4+n^3+n^2+n+1>n-1$. Hence $p$ cannot divide $n-1$, because otherwise $n-1$ would be at least $2^{m}p$, and $n^4+n^3+n^2+n+1$ would be at most $p$, which is less than $2^{m}p$. Hence, we have $$2^m+1=n,p^2=n^4+n^3+n^2+n+1.$$ --- ## Croatia2016_booklet — Page 18 FINAL ROUND – NATIONAL COMPETITION Let us notice that the second equation is equivalent to the following equations: $$n^4+n^3+n^2+n=p^2-1,$$ $$n(n+1)(n^2+1)=(p-1)(p+1).$$ By plugging $n=2^m+1$ into the last equation we get $$(2^m+1)(2^m+2)(2^{2m}+2^{m+1}+2)=(p-1)(p+1),$$ which leads us to