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The Sum of Fourth Powers (Posted on 2007-07-04) Difficulty: 3 of 5
Find all integer solutions to x4+y4=(x+y)(xy+1) with x≥y≥0.

See The Solution Submitted by Brian Smith    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 2

Substituting x=0 and y=0 in turn, having regard to the restriction
x> =y> = 0, we obtain:
(x, y) = (1, 0); (0, 0)

Now, let us consider the case x=y(=m, say).

Then, by the problem:
2*m^4 = (2m)*(1 + m^2)
Or, m(m^3 - m^2 - 1) = 0

So,
Either: m=0, giving x=y=0;
Or: m^3 - m^2 - 1 = 0, which do not yield any integer solutions.

For x> y> = 1; substituting (u, v) = (x+y, x-y); we obtain:
u> v> = 1, and:

x^4 - y^4< x^4 + y^4 = (x+y)(1+xy); giving:
v(u^2+v^2)/2< 1 + (u^2 -v^2)/4
or, 4+u^2-v^2> 2v(u^2+v^2)> = 2(u^2+v^2)
Or, u^2 + 3*v^2 < 4.

The above relationship cannot be satisfied for integers u and v with the restriction u> v> = 1.

This leads to a contradiction.

Consequently, (x, y) = (0, 0) and (1, 0) are the only possible solutions to the given problem.

 

Edited on July 4, 2007, 2:51 pm
  Posted by K Sengupta on 2007-07-04 14:49:31

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