Prove that $ 2^r +2$ is sum of 2 square number, when r is a prime number not equal to 2

Question

I was doing a question. Suddenly I got stuck at this last part of the problem. It was to prove $ 2^r +2 = a^2 +b^2$ where $r \neq 2$, r is a prime and $ a \neq b$. Also $r^2 -1$ is a mersenne prime. I tried to use fermat's little theorum, but to no avail. Thank you.

PS note: The problem I was solving was BMO2021 Q6.

Answer 1

$2^{2k+1}+2=(2^k-1)^2+(2^k+1)^2$.

Answer 2

The question is wrong. If $r=11$, $2^{11}-7 = 2041$ is not a square number.

Answer 3

Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
n = 2;
ParallelTable[
  If[TrueQ[2^r + 2 == a^2 + b^2 && a != b], {r, a, b}, Nothing], {r, 
   2, 10^n}, {a, 0, 10^n}, {b, 0, 10^n}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{3, 1, 3}}, {{3, 3, 1}}}, {{{5, 3, 5}}, {{5, 5, 3}}}, {{{7, 3, 
    11}}, {{7, 7, 9}}, {{7, 9, 7}}, {{7, 11, 3}}}, {{{9, 15, 
    17}}, {{9, 17, 15}}}, {{{11, 5, 45}}, {{11, 23, 39}}, {{11, 31, 
    33}}, {{11, 33, 31}}, {{11, 39, 23}}, {{11, 45, 5}}}, {{{13, 25, 
    87}}, {{13, 63, 65}}, {{13, 65, 63}}, {{13, 87, 25}}}}

So, we can see that you're not right! Because, for example, when $\text{r}=9\notin\mathbb{P}$ we get $2^9+2=15^2+17^2$.