Solving a Diophantine equation|Use a simple trick to find the roots

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  1. ShakitovMath
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Not so hard but a interesting problem. I like problems like this!

chenhanwen
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Awesome video as always🙂 thanks so much

SuperYoonHo
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Completing the square looks like overkill to me. Just throw the y^2 to the other side and notice that x^2022 is never negative and almost always HUGE while 2y-y^2 is almost always negative and never big when it's positive, which means X can only be -1, 0 or 1. Solving for these 3 cases is trivial.

akaRicoSanchez
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The equation is also equal to x^2=y(2-y). Hence y=0, y=1 or y=2.

jagtmrx
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Great video however this is a very simple case of completing the square with respect to y. More complex variations of this problem will require treating one of the variables as a constant and using the Quadratic formula and then setting the discriminant equal to a perfect square in order to obtain integer solutions.

moeberry
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y^2-2y+x^2022 = 0, therefore y = 1 ± √ 1 - x^2022 then easy to get all the answers

johnfeng