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.
Ok.
Solution is also:
n = (u + v)^(1/3) - 2/[9(u + v)^(1/3)] - 1/3
where:
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...
You can factorise it as The factor clearly cannot be a perfect square. So at least one of the squared prime factors in must be shared between the two factors on the left side. But the only possible common factor of and is 2. So we can write and for some natural numbers
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.