Empty one year ago Prove

1. Empty

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.

2. Empty

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

3. amilapsn

use $a^2+b^2=(a+b)^2-2ab$

4. Empty

Yep that'll do it. :D

5. amilapsn

:D

6. ganeshie8

Clever!

7. Empty

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

8. ganeshie8

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

9. Empty

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

10. ganeshie8

little challenge : prove that $$5$$ is a divisor whenever $$5\nmid n$$

11. Empty

Oooh interesting I'm thinking.

12. ganeshie8

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

13. Empty

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

14. ganeshie8

let $$a=2mx^2$$ and $$b=my^2$$ and let $$5\nmid n$$, then : $a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}$

15. ganeshie8

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

16. ganeshie8

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

17. ganeshie8

Here is the revised challenge (not a challenge anymore) : prove that $$5$$ is a divisor whenever $$5\nmid n,a,b$$

18. ganeshie8

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