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 Powers of x and y (Posted on 2017-01-21)
Find all possible couples of distinct integers (x,y) satisfying the equation: x^4 + y = x^3 + y^2

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 Another approach Comment 3 of 3 |

Rearranging gives:    x3(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   x3 = y – 1  and  x – 1 = y

(which is not possible since x3 – x + 2 = 0 has no integer roots)

or         x3 = 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 2017-01-22 10:57:30

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