Each of x and y is a positive integer such that x² + y² + x is divisible by 2xy.

Prove that x is the square of an integer.

(In reply to re: Do any exist? by xdog)

Gamer, this is why I suggested to KS that '2xy' might be a challenge, as I thought there might be a misprint.

I believe the correct answer is that this cannot be proved, since there are no qualifying squares, x. 

I did some workings to show that if x² + y² + x is divisible by 3xy, there are solutions but can't find my workings at the moment. I'll post again if I find them.

Edited on March 10, 2011, 2:50 am

I was only able to find a fragment of my working (I don't guarantee it's right, as it was just to support my comment to KS):

Assume that (x^2+x+y^2)/(2xy)=a
Then x^2+x+y^2=2axy
x(x+1)=y(2ax-y)
small values:
a=sqrt(3/2),x=2,y= sqrt(6)
a=2/sqrt(3),x=3,y=2sqrt(3)
a=sqrt(5)/2,x=4,y=2sqrt(5)
a=sqrt(6/5),x=5,y=sqrt(30)

g.f. a=((x+1)^(1/2))/(x^(1/2)), y=2*((x+1)^(1/2))
To clear the square roots we need some number x such that x is a square and (x+1) is n (with n a square) times another square, i.e a^2=n^2b^2-1; but no two squares differ by one.

However, if 3xy is substituted for 2xy, if (x^2+x+y^2)/(3xy)=a=1, then x is a square {1,4,25,169..}

 

Edited on March 10, 2011, 5:20 am

Posted by broll on 2011-03-10 02:49:33