Should I use the Division Algorithm to solve this or investigate 4 cases in which either x or y is even/odd or they are both even and odd? I don't think my instructor would accept the latter. Thank you.
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2Try modulo $10$: no square number ends in $2$ or $7$. – Michael Hoppe Apr 13 '20 at 18:40
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Welcome to Mathematics Stack Exchange. $2$ is not a quadratic residue modulo $5$ because $5\not\equiv\pm1\pmod8$ – J. W. Tanner Apr 14 '20 at 01:17
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This is most easily proven using mods. We know that if such a pair exists, then $x^2\equiv2\bmod5$.
Note that $$0^2\bmod5=0$$$$1^2\bmod5=1$$$$2^2\bmod5=4$$$$3^2\bmod5=4$$$$4^2\bmod5=1$$
So this is not possible.
Rushabh Mehta
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Yes that makes it easier. I should have thought about this. Thank you, Don! – Uyen Le Apr 13 '20 at 19:58
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$x=5k, 5k\pm1, $ or $5k\pm2,$
so $x^2=25k=5(5k), $
$x^2=25k^2\pm10k+1=5(5k^2\pm2k)+1$,
or $x^2=25k^2\pm20k+4=5(5k^2\pm4k)+4.$
In no case is $x^2=5y+2$.
J. W. Tanner
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