Name: sarathy

Solve :

1! + 2! + 3! + ... + x! = y ²

How do i find the solutions ?

There cannot be that many; can you find the small ones?

Now suppose that x is big, and look at the remainder of the division of 1! + 2! + 3! + 4! + 5! + ... + x! by 5.
Compare this with the possible remainders of divisions of squares by 5.

What does this tell you?