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

Note by Sharky Kesa 2 years, 1 month ago

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If \(n\) 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 \(5\mid n\).

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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.

You could just post it directly to me on Slack.

Bhai, can you come on Slack ?

Sure. Give me a few moments.

@Satyajit Mohanty – Of Course!

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## Comments

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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 \(5\mid n\).

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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.

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You could just post it directly to me on Slack.

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Bhai, can you come on Slack ?

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Sure. Give me a few moments.

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