The numbers x,y are chosen from {1,2,3........5N}.The probability that x^4-y^4 is divisible by 5 is?

Question

anonymous · Answer

Find the generalised  result i.e for all possible N

anonymous · Answer

$$x,y \in \mathbb{N} \:\: ?$$

anonymous · Answer

Yep

anonymous · Answer

This is tricky, and considering the fact I suck at probability, it gets even more tricky for me. 
$$x^4-y^4 = (x-y)(x+y)(x^2+y^2) $$
Now I need to get solutions for each factor.

anonymous · Answer

How to do that?

anonymous · Answer

$$x^4 + y^4 = 5k \implies (x-y)(x+y)(x^2+y^2)=5k$$ I'll ping @FoolForMath and @Mr.Math  , maybe they could do it... I will need to think about it, it will take time for me.

dominusscholae · Answer

From what I can think, this problem says this: what is the probability that either x-y, x+y, or x^2+y^2 is a multiple of 5. for x^2+y^2, 3^2+4^2=5^2, so this leads me to believe that the two numbers picked in this case make this part of problem "what is the probability that two numbers picked in N with one number being a multiple of 3 and the other a multiple of 4. The only problem here are the other two cases...

nikvist · Answer

Can x^4-y^4 be negative?

anonymous · Answer

Yeah, I think so...

nikvist · Answer

$$17/25=0,68$$

mr.math · Answer

That's very smart nikvist! You're the hidden great mind in this site :)

anonymous · Answer

Solution comes as a function of N

nikvist · Answer

I think, solution does not depend on N.