Question

Prove

Answer 1

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.

Answer 2

I guess this is either going to be hard or easy I'm not sure which it is cause I know the answer lol

Answer 3

use \[a^2+b^2=(a+b)^2-2ab\]

Answer 4

Yep that'll do it. :D

Answer 5

:D

Answer 6

Clever!

Answer 7

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

Answer 8

spoiler :

\(\color{white}{\sqrt{2ab} \in \mathbb{Z}\implies ab=2c^2 \\~\\a^2n^4+b^2 = (an^2)^2+b^2 = (an^2+b)^2-2an^2b = (an^2+b)^2-(2cn)^2}\)

Answer 9

Yeah that's exactly how I did, it, with c and everything haha.

Answer 10

little challenge :

prove that \(5\) is a divisor whenever \(5\nmid n\)

Answer 11

Oooh interesting I'm thinking.

Answer 12

sry that is a blunder, doesn't work...

Answer 13

I am still interested in how you thought it would have worked though :D

Answer 14

let \(a=2mx^2\) and \(b=my^2\) and let \(5\nmid n\), then :

\[a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}\]

Answer 15

similarly a fake proof can be made for the other case :
\(a=mx^2\) and \(b=2my^2\)

Answer 16

btw \(m\) is the square free factor, that must match in both \(a\) and \(b\) for \(2ab\) to be a perfect square

Answer 17

Here is the revised challenge (not a challenge anymore) :

prove that \(5\) is a divisor whenever \(5\nmid n,a,b\)

Answer 18

i lost interest immediately because of the constraint, \(5\nmid n,a,b\)