 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Cubic Difference Conclusion (Posted on 2015-02-21) Find all possible integer solutions to this equation:
P3 - Q3 = 2*P*Q + 8

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1

P3 – Q3 = 2PQ + 8                                               (1)

Changing to new integer variables that can be separated:

Let        P + Q = a    and   P – Q = b        so that

PQ  = (a2 – b2)/4

and       P2 + Q2 = (P + Q)2 – 2PQ = a2 – (a2 – b2)/2

(1) becomes  (P – Q)(P2 + Q2 + PQ) = 2PQ + 8

b(a2 – (a2 – b2)/4) = (a2 – b2)/2 + 8

3a2b + b3 = 2a2 – 2b2 + 32

Thus     a2 = (32 – 2b2 – b3)/(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, a2 = 29, so no integer solution.

When b = 2, a2 = 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 2015-02-23 08:11:02 Please log in:
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