P^{3
}– Q^{3} = 2PQ + 8 (1)
Changing to new integer variables that can be separated:
Let P + Q = a and
P – Q = b so that
PQ = (a^{2} – b^{2})/4
and P^{2}
+ Q^{2} = (P + Q)^{2} – 2PQ = a^{2} – (a^{2} –
b^{2})/2
(1) becomes (P – Q)(P^{2} + Q^{2}
+ PQ) = 2PQ + 8
b(a^{2} – (a^{2}
– b^{2})/4) = (a^{2} – b^{2})/2 + 8
3a^{2}b +
b^{3} = 2a^{2} – 2b^{2} + 32
Thus a^{2} = (32 – 2b^{2}
– b^{3})/(3b – 2) (2)
The cubic numerator of (2) has positive stationary
values when b = 0 and b = 4/3, and is zero between
b = 2 and b = 3. So it is positive for integer b <= 2.
The denominator of (2) is positive for integer b >=1.
So a is real only when integer b = 1 or 2.
When b = 1, a^{2} = 29, so no integer solution.
When b = 2, a^{2} = 4, giving a
= +2 or –2.
So only two solutions:
(a, b) = (2, 2) gives
(P, Q) = (2, 0)
and (a, b) = (2, 2) gives (P, Q) = (0, 2).

Posted by Harry
on 20150223 08:11:02 