/sci/ - Science & Math

>>7497923

First of all, non-triviality of any of our triples is easily dispensed with in this case, since for any natural q, gcd(q,q+1) (or as Nyblom might write, simply (q,q+1) ) is equal to one (left as an exercise), and thus bringing any natural hypotenuse into the mix doesn't matter, a given AIRA triple has relatively prime components, and each one stands very much on its own (of course, the multiples wouldn't even be AIRA triples!). So what's left is to show infinitude, and the if-and-only-ifness of the thing. But first, we must relate square triangular numbers to AIRA triples in the first place, and Nyblom does this in the following way. The letters are (generally) natural numbers.

- m and its successor, squared, require some s (basis of some odd number).

- m and s necessarily satisfy certain inequalities, a series of expressions constrain m and s, requiring some r to exist

- a quadratic in s (taking r along for the ride) is solved, and the discriminant has to be such-and-such. The negative radical has to be thrown out (left as an exercise)

- The discriminant involving r requires still further that some n exist (basis of another odd number), phew, it's getting tedious by now, we're in the tall weeds. But this is the most delightful part of Nyblom's note --- :^) ---

- rearrangement of r and n is precisely A SQUARE TRIANGULAR NUMBER

<span class="math"> \displaystyle r^2 = \frac{n(n+1)}{2} [/spoiler]

in its two aspects, of which there are infinitely many per the lemma! Either side IS the square triangular number. So the whole thing gets walked back right up the track, establishing the formula, and infinitude of solutions!