Find all possible couples of distinct integers (x,y) satisfying the equation:
x^4 + y = x^3 + y^2
Rearranging
gives: x^{3}(x – 1) = y(y – 1) (1)
If both sides are 0, we readily obtain the (x, y) pairs:
(0, 0), (0, 1), (1, 0), (1,
1).
If both sides are 1 or 1, clearly no possible integer pairs exist.
Any other possibilities can only arise from each side having at
least one prime factor, and since the consecutive integers x with
x – 1 and y with y – 1 can have no common prime factors, it
follows that
either x^{3} = y – 1 and x – 1 = y
(which is not possible since x^{3}
– x + 2 = 0 has no integer roots)
or x^{3}
= y and x – 1 = y – 1 giving x = 0, 1 or 1.
x = 0 and x = 1 do not give extra possible pairs, but, for x = 1,
each side of (1) has the value 2 and this gives two more pairs:
(1, 1) and (1, 2).
giving the six pairs posted earlier by Brian.

Posted by Harry
on 20170122 10:57:30 