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# Number Theory #4

Let $$n$$ be a positive integer such that $$2n+1$$ and $$3n+1$$ are both squares. Prove that $$40|n$$.

Note by Victor Loh
3 years, 4 months ago

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It suffices to prove $$5\mid n,8\mid n$$.

Given that $$2n+1=x^2, 3n+1=y^2$$, If we add the two equations we get: $$5n+2=x^2+y^2\Rightarrow x^2+y^2\equiv 2\pmod 5$$, since the quadratic residues of mod $$5$$ are$$0,1,4$$, we can check that only $$x^2\equiv y^2\equiv 1$$ works, therefore $$2n\equiv 1-1\equiv 0\pmod 5\Rightarrow 5\mid n$$

The same analysis can be done for $$8$$, but we will multiply the second equation by $$2$$ first and then add: $$8n+3=x^2+2y^2\Rightarrow x^2+2y^2\equiv 3\pmod 8$$, since the quadratic residues of mod $$8$$ are $$0,1,4$$, we can check that only $$x^2\equiv y^2\equiv 1$$ works again, which means $$3n\equiv 1-1\equiv 0\pmod 8\Rightarrow 8\mod n$$ and we are done.

- 3 years, 3 months ago

Great!

- 3 years, 3 months ago