obviously $k<3$. If $k\ge 3$ then $n^2\equiv 5 \pmod{8} $ and contradiction.

$4\mid n^2-1\implies 8\mid n^2-1\implies 2\mid (1+9\cdot 2^{k-2}+2^{2k-1})$ but the expression on the right is odd if $k\ge 3$. Then $k=1,2$ it remains to check them. Is this correct? Because it is way too easy.

If $ k \ge 3 \Rightarrow n^2 \equiv 5 \pmod {8} $, which is impossible since $ n^2 \equiv 0,1,4 \pmod {8} \forall n\in\mathbb{N} $ else, $ k=1 \Rightarrow n^2=31$, which is impossible and $ k=2 \Rightarrow n^2=73$, which is impossible, too! In conclusion, there is no solution. Ok, I forgot to check the case: $ k=0 $~So, the only solution is $ (k,n)=(0,4) $

BSJL wrote: If $ k \ge 3 \Rightarrow n^2 \equiv 5 \pmod {8} $, which is impossible since $ n^2 \equiv 0,1,4 \pmod {8} \forall n\in\mathbb{N} $ else, $ k=1 \Rightarrow n^2=31$, which is impossible and $ k=2 \Rightarrow n^2=73$, which is impossible, too! In conclusion, there is no solution. What about $k=0$ ? Thats the only solution $k=0$ and $n=4$.

assume 2^k=a, and then factorise the LHS to get a standard diophantine eq

Olemissmath wrote: Find all non-negative integers $k,n$ which satisfy $2^{2k+1} + 9\cdot 2^k+5=n^2$. cant we solve this using quadratics?

Olemissmath wrote: Find all non-negative integers $k,n$ which satisfy $2^{2k+1} + 9\cdot 2^k+5=n^2$.