Considera the diophantine equation:
$y^2=x^2+(x+449)^2$.
Is there a method to solve this equation?
And why an Oeis sequence Is dedicated to this equation? Has this diophantine equation something special, has this diophantine equation some context?
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1https://oeis.org/A159589 gives a recurrence for $a(n)$ - so, yes, this looks like the full solution. – Dima Pasechnik Jul 30 '23 at 20:59
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@Twiga Welcome to MathOverflow. FYI, using an Approach0 search, the Math SE site's Can OEIS be used for searching of sequences of pairs? has an answer that states "... the solutions $(x,y)$ of $x^2+(x+31)^2=y^2$ are A118674, A157646". Thus, there is at least one other similar Diophantine equation in OEIS, but there's also no reason given for why it's there. – John Omielan Jul 30 '23 at 22:04
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@Twiga As for solving it, $449$ is prime so there are $2$ basic cases. With $\gcd(x,449)=d$, then if $d=1$, we can leave your equation alone, else $d=449$ so we can set $x=449a$ & $y=449b$ to get $b^2=a^2+(a+1)^2$. Then we're trying to find primitive Pythagorean triples, using the formula provided there, but also handling the possibility that $x$ and/or $x+449$ are negative. Solving the equations will then result in a generalized Pell's equation. – John Omielan Jul 30 '23 at 22:13