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This question is an offshoot of Ratio of triangular numbers. Suppose $ka(a+1)=nb(b+1)$, where $k,n >1$ are relative prime integers, and $a,b \geq 0$ are integers. Which $k,n$ pairs have no solution other than the trivial one $a = 0, b = 0$?

(Checked for $k,n<21$. Found no solutions for $4,9$; $9,16$; and - surprise - $16,19$. The rest have solutions in well-defined families. E.g. for $k+1=n$:

$a(0)=0; a(1)=2k+1; a(n)=(4k+2)a(n-1)-a(n-2)+2k;$

$b(0)=0; b(1)=2k; b(n)=(4k+2)b(n-1)-b(n-2)+2k$.)

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  • $\begingroup$ Perhaps you can give some motivation to this question. $\endgroup$ – Chris Ramsey Nov 24 '15 at 18:11
  • $\begingroup$ $(16)(6981194415)(6981194415+1) = (19)(6406383360)(6406383360+1)$ $\endgroup$ – Aeryk Nov 24 '15 at 18:50
  • $\begingroup$ Thank you, Aeryk. I thought so, but I couldn't go that high. $\endgroup$ – Adam Kertesz Nov 24 '15 at 19:15
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There should always be solutions unless $kn$ is a square. The equation is equivalent to $$k (2a+1)^2 - n (2b+1)^2 = k - n.$$ Let $(x_0, y_0)$ be the fundamental solution of the Pell equation $x^2 - 4 k n y^2 = 1$. Then $$ a = \frac{x_0-1}{2} + n y_0, \quad b = \frac{x_0-1}{2} + k y_0 $$ give you a solution.

For $k = 16$, $n = 19$, a solution is $$a = 6981194415, \qquad b = 6406383360.$$ (See also Aeryk's comment to the question.)

Now assume that $kn$ is a square. We can obviously reduce to the case that $k$ and $n$ are coprime. Then $k = k_1^2$ and $n = n_1^2$, and we have the equation $$ (k_1 (2a + 1))^2 - (n_1 (2b+1))^2 = k_1^2 - n_1^2. $$ The left hand side factors as $(k_1(2a+1) + n_1(2b+1))(k_1(2a+1) - n_1(2b+1))$. If $k_1^2 - n_1^2 = rs$ is the corresponding factorization of the right hand side, then we get $$2a+1 = \frac{r+s}{2k_1} \qquad\text{and}\qquad 2b+1 = \frac{r-s}{2n_1}$$ (assuming $r \ge s$). $r = k_1 + n_1$, $s = k_1 - n_1$ gives the trivial solution (when $k_1 > n_1$). There can be other factorizations that work, but there may be none. For example, with $k = 6^2$, $n = 1$, we have $6^2 - 1 = 35 \cdot 1$, which gives $a = 1$, $b = 8$. In any case, there are only finitely many solutions, whereas in the nonsquare case, there will always be infinitely many.

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  • $\begingroup$ Thank you very much, @Michael Stoll. That's it. Now I can sleep. $\endgroup$ – Adam Kertesz Nov 24 '15 at 19:27

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