Chavi's Mathematical adventure (+Karma for solutions)

[brightsky]

I hate this solution, if you may call it that, but it's probably the fastest way to do Q5:

So sub a = 2x and b = 2y in:
4(x+y)^2 = 50x...[1]
36x^4 + 2y^4 = 41xy^2...[2]

Look at equation [1]:
Since x, y are both positive integers, then 50x must be a multiple of 4, which means x = 2q, where q is some integer.
4(2q + y)^2 = 100q
2q + y = 5sqrt(x)
Again, since q and y are both positive integers, we need x = p^2, where p is some integer.
So we have 2q = p^2
The only solution here (besides 0) is q = 2 and p = 2, which means x = 4, which in turn means that y = 6.

[pi]

That brilliant!

Thanks brightsky (+1 karma well deserved there)

[pi]

I think this is what you meant brightky (you got x=4, which would make a=8... which is incorrect - would also make b = 12)



Using



As and ,
As , let
As , let

or
As ,





Still have no idea on question 6...