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Show that : $$\forall \hspace{2mm} (k,j,m,n) \in \mathbb{N}^4 : 3kn^2 +3jn +3m+2 \notin S=\{x^2 \mid x\in \mathbb{N}\}$$ Clearly this is solvable using modular arithmetic because claiming the expression is a perfect square would yield the congruence $z^2 \equiv 2 \pmod 3$ which of course is a contradiction.

But I'm intrigued if there is a way of proving it without having to use modular arithmetic , perhaps in a more elementary way .

Thomas Andrews
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Adam Boussif
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1 Answers1

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Move the LHS $2$ to the RHS, we get a multiple of $3$ on LHS.

RHS is $x^2-2 = x^2-1-1 = (x-1)(x+1)-1$

If $x$ is $1$ more than or $1$ less than a multiple of $3$ , then this RHS $(x-1)(x+1)-1$ is not a multiple of $3$ , & thus not Equal to LHS.

If $x$ is a multiple of $3$ , then this RHS $x^2-2$ is not a multiple of $3$ , & thus not Equal to LHS.

Overall, it is not Possible.

[[ This is Modular Arithmetic in Disguise !! There is no way to escape that !! ]]

Prem
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    oh ! i see that's very clever i've thought about the same answer but wasn't quite satisfied yet , but now my doubts are cleared so thank you ! – Adam Boussif Jul 24 '22 at 17:33