Equation $$a^2=2^n-k$$ $$a,n,k>1 \quad \text{and} \quad k<n$$ has only 1 solution: $a=181$. How to prove that there are no more solutions for greater $a$?
What I have found:
- $k$ is not multiple of 3 and 5; and $k=7 \bmod 8$.
- There are no solutions for even $n$. Roughly: $d<d+a<(d-a)(d+a)=k<n$, where $d=2^{n/2}$.
- There are no solutions for odd $n$ if $a\neq [b\sqrt 2]$, where $2b^2=2^n$ and $[\cdot]$ is floor (the integer part). The proof is similar to the above.
- $[b\sqrt 2]\{b\sqrt 2\}$<n/2, where $\{\cdot\}$ is fractional part. Irrationality and ergodicity imply that for big $n$ the probability of this inequality is $P\approx n/[b\sqrt 2]$. Numerically checking the first $10^4$ numbers, the probability that another solution exists is $P=10^{-1498}$, which is practically $0$.
The question again: is there a way to prove the claim exactly?
