x⁴ + (x+1)⁴ = y² + (y+1)²

Source: http://mathforum.org/wagon/spring03/p978.html

Solution.

When you see the solution, you'll say "how did anyone ever think of that?" so I'll tell you how I really solved it.  First, I used an Excel spreadsheet to make a table of the first hundred or so values of the left side of the equation, and the first hundred or so values of the right side.  I didn't find any matches except x=y=0.  So I surmised that the trick is to show there are no other matches.  I looked for patterns in modulo arithmetic (patterns like the left-hand-side is always 1 mod 8 while the right-hand-side is always 5 mod 8), checking bases from 2 to 25, and found some interesting patterns, but nothing glaringly helpful.

Then, I noticed that the left side is 2x⁴ plus some lower powers of x, so that when x gets very large, the fourth power predominates.  If I double the left hand side, then take the square root, the result should be close to an integer.  I used Excel to make  a table of the first fifty or so values of

sqrt( 2x⁴ + 2(x+1)⁴ )

and found this gets very close to the square of an even integer as x increases.  Similarly, 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.

Using the method of successive differences, I pieced together the polynomial representing the sequence of square roots of twice the left side, and the sequence of square roots of twice the right side.

Left side: sqrt( 2x⁴ + 2(x+1)⁴ ) gets closer and closer to an integer which is given by 2x²+2x+2

Right side: sqrt( y² + (y+1)² ) gets closer and closer to an integer which is given by 2y+1

Twice the left side turns out to be (2x²+2x+2)²-2, and twice the right side turns out to be (2y+1)²+1

So the left side is 2 less than a perfect square, and the right side is 1 more than a perfect square, and since they're equal, we have the difference of two squares being 3, which means the two squares in question are 1 and 4.

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 it the only solution.