Recently I came to know from a friend that the only non-trivial value of satisfying the Diophantine equation

is . And apparently, although the problem looks fairly simple, especially since we have the formula for the left hand side

the proof is hard. In fact, I was told that all proofs of this result use advanced machinery. Nevertheless, given the simple nature of the problem I was intrigued to give it a try.

So let us assume that

.

Noting that , we have that and are coprime. Indeed, if is a common factor, then has to divide both and , forcing . So we deduce that

or

for some integers .

Note that so . Hence the possibilities are:

(i) , .

(ii) , .

First suppose that . Then is a square triangular number, so

for some , where are the Pell numbers. and are the half-companion Pell numbers, and they are given by for each . So we have to find all such that

.

On the other hand, if , then for some integer . Then . Hence implies

.

Now , so for some integers . Clearly , so

and .

So is a near isosceles Pythagorean triple, i.e.

.

Hence

.

The two boxed cases are what only remain to check, and they show why this actually is a difficult problem! (Refer to the articles about Fibonacci numbers in the ‘Notes and Articles’ tab above to see how little information we have on divisors of sequences of this kind.) Nevertheless, I’ll leave it here and maybe come back to it another time.

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