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Partial solution:It can be shown that when $5n+1$ and $7n+1$ (where $n\in\mathbb{N}$) are both perfect squares, then $n$ is divisible by $24$.
Find a method for generating all such $n$.
Good work, Opalg!Partial solution:If $5n+1=p^2$ and $7n+1=q^2$ then $35n+7 = 7p^2$ and $35n+5 = 5q^2$. Subtract, to get $7p^2 - 5q^2 = 2$. That is a Diophantine equation, which you can solve by looking at the continued fraction expansion of $\sqrt{5/7}$. If the $k$th solution is $(p_k,q_k)$, then the first few solutions, with the corresponding values $n_k$ of $n$ are $$\begin{array}{c|c|c|c}p_k&q_k&n_k&n_k/24 \\ \hline 1&1&0&0 \\ 11&13&24&1 \\ 131 & 155 & 3432 & 143 \\ 1561 & 1847 & 487344 & 20306 \\18601 & 22009 & 69199440 & 2883310 \end{array}$$
The numbers $p_k$ satisfy the recurrence relation $p_k = 12p_{k-1} - p_{k-2}$, and $q_k$ satisfies the same recurrence relation. That enables you to generate all the values of $n$, but I have not been able to find an explicit formula for $n_k.$
Difference equations...they're comin' for ya...And this is yet another example of why I am an engineering major...
Sorry for interrupting with a somewhat "useless" post. Carry on.
Weren't you posting problems with differential equations?And this is yet another example of why I am an engineering major...
Sorry for interrupting with a somewhat "useless" post. Carry on.