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Question Number 84461 by Tony Lin last updated on 13/Mar/20

$$p^2+3q^2=11907,p,q\in \mathbb{Z},findp\mathrm{\&}q$$

Commented bymr W last updated on 13/Mar/20

$$igotonlytwosolutions:$$ $$p=0,q=\pm 63$$ $$p=\pm 108,q=\pm 9$$

Commented byTony Lin last updated on 13/Mar/20

$$p^2+3q^2=3^5\times 7^2$$ $$\Rightarrow p^2=3(63+q)(63-q),q⩽63$$ $$\Rightarrow p^2=3^5(7+\frac{q}{9})(7-\frac{q}{9})$$ $$\because p\mathrm{\&}q\in \mathbb{Z}$$ $$\therefore 9\mid q$$ $$\to checkthat(7+\frac{q}{9})(7-\frac{q}{9})=3\times n^2,n\in \mathbb{Z}$$ $$(7+1)(7-1)=3\times 4^2$$ $$(7+7)(7-7)=3\times 0^2$$ $$\Rightarrow \{\begin{array}{l}p=0,q=\pm 63\\ p=\pm 108,q=\pm 9\end{array}$$

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