Find all integral pairs (x,y) that satisfy
x^3 - y^3 = xy + 61
thanx in advance
It looks as though there are only two solutions.
Factorise the left side: $\displaystyle (x-y)(x^2+xy+y^2) = xy + 61.\qquad(1)$
Let $\displaystyle k=x-y$. Then (1) becomes
$\displaystyle kx^2 + (k-1)xy + ky^2 = 61. \qquad (2)$
Complete the square to write this as
$\displaystyle k\Bigl(\bigl(x+\tfrac{k-1}{2k}y\bigr)^2 + \bigl(1-\tfrac{(k-1)^2}{4k^2}\bigr)y^2\Bigr) = 61.\qquad(3)$
Since k is a nonzero integer, $\displaystyle (k-1)^2\leqslant 4k^2$, and so the large parenthesis on the left side of (3) must be positive. Hence so also is k, or in other words $\displaystyle x>y$.
Thus if x is negative then so is y. But if (x,y) is a solution of (1) then so is (–y,–x). It follows that we only need to look for solutions where x is positive. Write the original equation as
$\displaystyle x^3 = y^3+xy+61.\qquad(4)$
It is clear from (4) that if x is positive then the only negative values that y can take are –1, –2 or –3. But it is easily checked that none of these leads to an integer solution of (4). We are left with the situation where each of the integers x, y and k in (2) is positive. It is then easily verified that the only way for the terms on the left of (2) to add up to 61 is if x=6 and y=5.
Therefore the only solutions to the original equation are (x,y) = (6,5) and (x,y) = (–5,–6).