Timeline for Does the equation $241+2^{2s+1}=m^2$ have a solution?
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| when toggle format | what | by | license | comment | |
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| Sep 28, 2015 at 19:32 | comment | added | R.P. | If you're looking for a congruence argument, then numbers of the form $M=2^{2n}-1$ are a smart choice of modulus, because we expect $241+2^{2s+1}$ to take only about $n$ distinct residue classes modulo $M$. So with that idea in hand, I guess you could do a computer search that runs through some small $n$ and checks which one gives an $M$ that does the job. | |
| Sep 28, 2015 at 7:28 | vote | accept | few_reps | ||
| Sep 28, 2015 at 7:12 | comment | added | few_reps | Thanks ! How did you think to 63 ? Will's answer below seems to indicate that obstructions are to be found mod $2^{2t}-1$ ... | |
| Sep 27, 2015 at 22:32 | history | answered | Stefan Kohl♦ | CC BY-SA 3.0 |