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Question Number 120923 by bemath last updated on 04/Nov/20

$$\mathrm{Determine}\mathrm{all}\mathrm{primes}p\mathrm{for}\mathrm{which}$$$$\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}$$$$\{\begin{array}{l}p+1={2\mathrm{x}}^2\\ p^2+1={2\mathrm{y}}^2\end{array};\mathrm{has}\mathrm{solution}\mathrm{in}$$$$\mathrm{integers}\mathrm{x},\mathrm{y}.$$

Answered by liberty last updated on 04/Nov/20

$$\mathrm{Note}\mathrm{that}p+1=2x^2\mathrm{is}\mathrm{even}\mathrm{so}\ne 2\mathrm{also}$$$${2\mathrm{x}}^2\equiv 1\equiv {2\mathrm{y}}^2\left(\mathrm{mod}\mathrm{p}\right)\mathrm{which}\mathrm{implies}\mathrm{x}=\pm \mathrm{y}.\mathrm{Since}\mathrm{p}\mathrm{is}\mathrm{odd}$$$$\mathrm{and}\mathrm{x}<\mathrm{y}<p\mathrm{we}\mathrm{have}\mathrm{x}+\mathrm{y}=p,\mathrm{then}$$$$p^2+1=2{(p-x)}^2=2p^2-4px+p+1$$$$givep=4x-1;{2\mathrm{x}}^2=4\mathrm{x},\mathrm{x}\mathrm{is}0\mathrm{or}2.$$$$\mathrm{and}p\mathrm{is}-1\mathrm{or}7.\mathrm{Of}\mathrm{course}-1\mathrm{is}\mathrm{not}$$$$\mathrm{prim}\overline{\mathrm{e}},\mathrm{but}p=7\mathrm{is}\mathrm{prime}\mathrm{and}(\mathrm{x},\mathrm{y})=(2,5)\mathrm{is}\mathrm{a}\mathrm{solution}$$

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