There exists positive integers \(n\) such that \(2n + 1\) and \(3n + 1\) are perfect squares. Prove that \(n\) is divisible by \(40\).

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TopNewestIf \(n\) is odd, then \(2n+1=2(2k+1)+1=4k+3\). But \(3\) is not a quadratic residue mod \(4\).

If \(n\equiv \{2,4,6\}\pmod{8}\), then \(3n+1\equiv\{7,5,3\}\not\equiv x^2\pmod{8}\). Therefore \(8\mid n\).

If \(n\equiv\{1,3\}\pmod{5}\), then \(2n+1\equiv\{3,2\}\not\equiv x^2\pmod{5}\).

If \(n\equiv\{2,4\}\pmod{5}\), then \(3n+1\equiv\{2,3\}\not\equiv y^2\pmod{5}\). So \(5\mid n\). – Mathh Mathh · 1 year, 4 months ago

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Actually \(n\) is divisible by 40 as well. Try it. I've a solution, but I'm awaiting others to post theirs. Meanwhile, I'd say that the problem would have been much accurate had you asked us to prove that \(n\) is divisible by 40 instead of 8. – Satyajit Mohanty · 1 year, 4 months ago

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– Sharky Kesa · 1 year, 4 months ago

You could just post it directly to me on Slack.Log in to reply

– Swapnil Das · 1 year, 4 months ago

Bhai, can you come on Slack ?Log in to reply

– Satyajit Mohanty · 1 year, 4 months ago

Sure. Give me a few moments.Log in to reply

– Swapnil Das · 1 year, 4 months ago

Of Course!Log in to reply