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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.

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  • $\begingroup$ As it is a part of a question, it 'can' be incomplete in data. But most probably it is not. If you feel something more should have been provided, please point me out. $\endgroup$ Dec 13, 2020 at 14:34
  • $\begingroup$ So why not provide as much relating details as possible? In most cases it helps others to help you with the problem you have $\endgroup$
    – Divide1918
    Dec 13, 2020 at 14:36

3 Answers 3

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

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  • $\begingroup$ Ok. I was actually doubtful if it will be proved by doing so. $\endgroup$ Dec 13, 2020 at 15:22
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The question is wrong. If $r=11$, $2^{11}-7 = 2041$ is not a square number.

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  • $\begingroup$ Thank you for pointing it. I have edited the question. $\endgroup$ Dec 13, 2020 at 14:43
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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 $2^9+2=15^2+17^2$.

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  • $\begingroup$ See, I never said that it is going to be false for composite numbers. It can also be true for some composite numbers. $\endgroup$ Dec 13, 2020 at 14:53

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