Determine all pairs (x,y) of positive integers that satisfy this equation:
x3+y3= x2+42xy+y2
Suppose y = k*x ie y is a multiple of x
x^3 - x^2 - 42k*x^2 - k^2*x^2 + k^3*x^3 = 0
(k^3 + 1)x^3 - (k^2 + 42k + 1)x^2 = 0
x^2 ( (k^3 + 1)x - (k^2 + 42k + 1) ) = 0
(k^3 + 1)x - (k^2 + 42k + 1) = 0
x = (k^2 + 42k + 1) / (k^3 + 1) and this must be an integer
Plugging in 1 for k gives x=22. So if y = kx and k=1 then x=22.
Plugging in 7 for k gives x=1. So if y = kx and k=7 then x=1 and y=7.
Plugging in 1/7 for k gives x=7, y=1.
But I still have not proved these are the only solutions.
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Posted by Larry
on 2024-07-01 09:50:48 |
A bit more analytic progress
