Is (0,1) not a valid solution?
Find all positive integer roots of the equation
My solution attempt:
Following Wolframalpha factoring, it can be rewritten as , then
and putting and for the
first positive integer root,
second positive integer root,
How can I prove that these are the only positive integer solutions? Thanks for any suggestions. Regards.
Solution is also:
n = (u + v)^(1/3) - 2/[9(u + v)^(1/3)] - 1/3
u = SQRT(27A^4 - 40A^2 + 16) / [2(3^(3/2)]
v = (27A^2 - 20) / 54
(0,1) , (1,2) and (7,20) seem to be the only integer solutions with n=>0.
I ran a looper with A from 1 to 999999: only above resulted.
So perhaps above equation can be used to devise the proof...
Therefore which, with a bit of reorganisation and after cancelling the factor 2, can be written as This shows that the set is a Pythagorean twin triple. There are infinitely many Pythagorean twin triples, but we want one in which the middle element is a perfect square. The triples (0,1,1) and (3,4,5) satisfy that condition, leading to the values n=1 and n=7. The next triples are (20,21,29), (119,120,169) and (696,697,985), but none of those has a perfect square as the middle element. After that, the numbers increase very rapidly, as you can see from the listing at A046090 - OEIS. My guess is that there are no more perfect squares in that sequence, but I have to admit that I don't begin to see how to prove that.