Prove

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Prove

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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For a,b integers greater than 1, if \(\sqrt{2ab} \in \mathbb{Z}\) then \(a^2n^4+b^2\) is not prime for n>1.
I guess this is either going to be hard or easy I'm not sure which it is cause I know the answer lol
use \[a^2+b^2=(a+b)^2-2ab\]

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Yep that'll do it. :D
:D
Clever!
I came up with this after reading this http://www.cut-the-knot.org/blue/McWarter.shtml to try to generalize this question at the end. :P
spoiler : \(\color{white}{\sqrt{2ab} \in \mathbb{Z}\implies ab=2c^2 \\~\\a^2n^4+b^2 = (an^2)^2+b^2 = (an^2+b)^2-2an^2b = (an^2+b)^2-(2cn)^2}\)
Yeah that's exactly how I did, it, with c and everything haha.
little challenge : prove that \(5\) is a divisor whenever \(5\nmid n\)
Oooh interesting I'm thinking.
sry that is a blunder, doesn't work...
I am still interested in how you thought it would have worked though :D
let \(a=2mx^2\) and \(b=my^2\) and let \(5\nmid n\), then : \[a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}\]
similarly a fake proof can be made for the other case : \(a=mx^2\) and \(b=2my^2\)
btw \(m\) is the square free factor, that must match in both \(a\) and \(b\) for \(2ab\) to be a perfect square
Here is the revised challenge (not a challenge anymore) : prove that \(5\) is a divisor whenever \(5\nmid n,a,b\)
i lost interest immediately because of the constraint, \(5\nmid n,a,b\)

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