if $N=1(mod 2)$ and $N\ge 3$ then $2^N-2N=2(mod 4)$ is not a perfect square. if $N=0(mod 2)$ and $N\ge 8$,assume that $N=2k$, then $(2^k-1)^2<2^N-2N=2^{2k}-4k<(2^k)^2$ is not a perfect square.

yefangzhou wrote: if $N=1(mod 2)$ and $N\ge 3$ then $2^N-2N=2(mod 4)$ is not a perfect square. if $N=0(mod 2)$ and $N\ge 8$,assume that $N=2k$, then $(2^k-1)^2<2^N-2N=2^{2k}-4k<(2^k)^2$ is not a perfect square. How would you prove $\left({2^k-1}\right)^2 < 2^{2k}-4k$ ? I used an induction proof at the competition.

derivative could help.

I proved it in this manner: $\left({2^k-1}\right)^2 < 2^{2k}-4k$ is equivalent to $2^{k+1} > 4k+1$ Base: $k=3$ $16 > 13$ Induction hypothesis: $2^{k+1} > 4k+1$ for all positive integers $k \leq 3$ Inductive step: We have to show: $2^{k+2} > 4k+5$ From the induction hypothesis we know: $2^{k+2} >8k+2$ If we can show: $8k+2 > 4k +5$ then $2^{k+2} > 4k+5$ must also be true. This is equal to $k> 3/4$ which is obviously true for $ k \geq 3$ Thus, the equation $2^{k+1} > 4k+1$ must be true for $k \geq 3$ Checking the remaining even cases yields the solution 0 for $N=2$, checking the case $N=1$ yields the solution 1. (the case $N=1$ has to be cheked because $2^N-2N$ is not an even number for $N=1$) Thus $N=1$ and $N=2$ are the only solutions.

@above maybe this is more common: Let $P(k): 2^{k+1}>4k+1: k\ge 3$. Easily $P (3) $ works so assume $P(k) $ works for some $k\ge 3$. $P^{-1}(k+1):4(k+1)+1=(4k+1)+4<2^{k+1}+4 <2^{k+1}+2^{k+1}=2^{k+2} $ So $P (k+1) $ is true and the induction on $P$ holds