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A Strange Number Theory proof?

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
1 year, 8 months 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 \(5\mid n\). Mathh Mathh · 1 year, 8 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, 8 months ago

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@Satyajit Mohanty You could just post it directly to me on Slack. Sharky Kesa · 1 year, 8 months ago

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@Satyajit Mohanty Bhai, can you come on Slack ? Swapnil Das · 1 year, 8 months ago

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@Swapnil Das Sure. Give me a few moments. Satyajit Mohanty · 1 year, 8 months ago

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@Satyajit Mohanty Of Course! Swapnil Das · 1 year, 8 months ago

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