Answer: x=y=0

Solution:

I'll tell you how I solved this problem. I started by using a spreadsheet to explore the values of the left-hand-side and the right-hand-side for various integers x and y, respectively. I noticed that twice the left side is very close to the square of an even number, and twice the right side is very close to the square of an odd number, so I set about finding the particular even and odd numbers in question. I hoped that perhaps I could derive an equation equating the square of an even number to the square of an odd number, and get a contradiction.

Let a represent the left side, and let b represent the right side.

2a = 2x^{4}+2(x+1)^{4}

= 4x^{4} + 8x^{3} + 12x^{2} + 8x + 2

= (2x^{2}+2x+2)^{2} -2

2b = 4y^{2} + 4y + 2

= (2y+1)^{2} + 1

As it turns out, the "odd" and "even" aspect of this problem isn't important. What is important is that the only number that is both 2 less than a square and 1 more than a square is 2, so twice the left side is 2, and twice the right side is 2, hence x=y=0.

Ron v.d. Burg proposed another approach, which I like better than my own:

(a) y^{2} + (y+1)^{2} is strictly increasing as a function of y for nonnegative y.

x^{4} + (x+1)^{4} > (x^{2}+x)^{2} + (x^{2}+x +1)^{2} for x > 0; (difference is 2x^{2}+2x)

Using result (a),

(1) y > x^{2}+x for x > 0.

x^{4} + (x+1)^{4} < (x^{2}+x+1)^{2 }+ (x^{2}+x+1 +1)^{2} for x > 0; (difference is 2x^{2}+2x+4)

Using result (a),

(2) y < x^{2}+x+1 for x > 0.

Combining results (1) and (2) shows us that there are no solutions for x > 0.

Choosing x = 0 gives y = 0 as the only nonnegative solution.