Let $q>5$ be an odd sum of two nonzero squares and consider the equation $$X^2-mkX+m\frac{qk+1}{4}=0$$ for some integers $m\ge 1$ and $7\le k\equiv 3\pmod{4}$ depending on $q$. The condition on $k$ of course is so that the fraction is an integer. After running some test, it seems like
The equation has integer solutions for some $m,k$ if and only if $q$ is not a square.
This equation comes from trying to write a particular continued fraction as a sum of two unitary fractions, but the background is not really important here. Does anyone have any idea of how to start attacking this problem? From the numerical tests, it doesn't look like there is an obvious way to explicitly compute $m,k$ from $q$; so (assuming the claim is true) I am expecting a non constructive approach or method.
We shall prove that the claim is correct, by setting $q$ as a square and then showing that an integral solution will lead to contradiction on $k$.
Overview
Let $p,q$ be primes dividing $k$ and $m$ respectively by an odd number of times. First suppose $2$ divides $m$ an even number of times. Let $$ \left[\frac{a}{p}\right] $$ denote the Legendre symbol. The approach is to form simultaneous quadratic reciprocity equations for all $p$ and $q$ (pairwise) using the original equation. Then using also the Law of quadratic reciprocity $$ \left[\frac{p}{q}\right]\left[\frac{q}{p}\right] = (-1)^{(p-1)(q-1)/4}, $$ solving all the equations will force an even number for primes $p\equiv 3\pmod 4$ dividing $k$. This in turn will force $k\equiv 1 \pmod 4$, contradicting the original assumption.
When $2$ divides $m$ by an odd number of times, we need to first show that $k\equiv 7\pmod 8$. Then later solving of the simultaneous equations will reveal that $k\not\equiv 7 \pmod 8$.
Proof. Solving for $X$, we have $$ X = \frac{1}{2} \left(k m \pm \sqrt{k^2 m^2-km q-m}\right) $$ Since $X\in\mathbb Z$, we conclude that $$k^2m^2-kmq-m=w^2$$ for some $w\in\mathbb Z$. Write $m = \sigma^2 v$ such that $v$ is squarefree, then $$ \sigma^2k^2v^2-kvq-v=(w/\sigma)^2 $$ Here $w/\sigma\in\mathbb Z$, else $(w/\sigma)^2\not\in\mathbb Z$ which would contradiction LHS being integral. Since $v$ is squarefree, this shows that $v$ divides $w/\sigma$, therefore $$ \sigma^2k^2v-kq-1= v(\frac{w}{\sigma v})^2 = vy^2 $$ Now setting $k=\rho^2 u$, $u$ squarefree, and by assumption $q$ is a square, we have $$ (\sigma\rho^2)^2u^2v - u(\rho \sqrt q)^2 - 1 = vy^2 \Longleftrightarrow z^2u^2v - ux^2-1 = vy^2 $$ which is what we want. Clearly $D=\gcd(u,v)=1$, otherwise taking $\pmod D$ gives us $$ -1 \equiv 0 \pmod D $$ This completes the proof.
$$ \tag*{$\square$} $$
Observe that since $u$ and $v$ are squarefree, they are product of distinct primes. For each prime $p$ we can take $\pmod p$ to get a quadratic reciprocity equation, which is first the half of our goal.
We first assume that $2$ divides $m$ and even number of times, so that $v$ is now odd. The other case is much more complex.
We fix a prime factorization notation since our next step requires taking mod each prime:
For the second half of our proof, we use the simultaneous quadratic reciprocity equations to obtain an equality relating $b$ and $d$.
We remark that $\gcd(u,v)=1$ from proposition 1, so that all the symbols here are $\pm 1$.
Proof. From the equation in proposition 1, we obtain $$ \begin{align} vy^2 &\equiv -1 \pmod{r_i}\\ vy^2 &\equiv -1 \pmod{s_i}\\ ux^2 &\equiv -1 \pmod{t_i}\\ ux^2 &\equiv -1 \pmod{w_i} \end{align} $$ For $r_i,t_i\equiv 1 \pmod 4$, $-1$ is a quadratic residue and hence $$ \begin{align} 1 &= \left[\frac{v}{r_i}\right]=\left(\prod_{j=1}^c\left[\frac{t_j}{r_i}\right]\right)\left(\prod_{k=1}^d\left[\frac{w_k}{r_i}\right]\right)\\ 1 &= \left[\frac{u}{t_i}\right]=\left(\prod_{j=1}^a\left[\frac{r_j}{t_i}\right]\right)\left(\prod_{k=1}^b\left[\frac{s_k}{t_i}\right]\right) \end{align} $$ For $s_i,w_i\equiv 3 \pmod 4$, $-1$ is not a quadratic residue and hence $$ \begin{align} -1 &= \left[\frac{v}{s_i}\right]=\left(\prod_{j=1}^c\left[\frac{t_j}{s_i}\right]\right)\left(\prod_{k=1}^d\left[\frac{w_k}{s_i}\right]\right)\\ -1 &= \left[\frac{u}{w_i}\right]=\left(\prod_{j=1}^a\left[\frac{r_j}{w_i}\right]\right)\left(\prod_{k=1}^b\left[\frac{s_k}{w_i}\right]\right) \end{align} $$ Taking the products over all $r_i,s_i,t_i,w_i$, we have $$ \begin{align} 1 &= \prod_{i=1}^a\left[\frac{v}{r_i}\right]=\prod_{i=1}^a\left\{\left(\prod_{j=1}^c\left[\frac{t_j}{r_i}\right]\right)\left(\prod_{k=1}^d\left[\frac{w_k}{r_i}\right]\right)\right\}\\ (-1)^b &= \prod_{i=1}^b\left[\frac{v}{s_i}\right]=\prod_{i=1}^b\left\{\left(\prod_{j=1}^c\left[\frac{t_j}{s_i}\right]\right)\left(\prod_{k=1}^d\left[\frac{w_k}{s_i}\right]\right)\right\}\\ 1 &= \prod_{i=1}^c\left[\frac{u}{t_i}\right]=\prod_{i=1}^c\left\{\left(\prod_{j=1}^a\left[\frac{r_j}{t_i}\right]\right)\left(\prod_{k=1}^b\left[\frac{s_k}{t_i}\right]\right)\right\}\\ (-1)^d &= \prod_{i=1}^d\left[\frac{u}{w_i}\right]=\prod_{i=1}^c\left\{\left(\prod_{j=1}^a\left[\frac{r_j}{w_i}\right]\right)\left(\prod_{k=1}^b\left[\frac{s_k}{w_i}\right]\right)\right\} \end{align} $$ Finally, we combine the 4 products into one, although we do not write it down. It can be observed that on the LHS we have $(-1)^{b+d}$, while the RHS can be observed to be the product of all combinations of $$ \left[\frac{r_i}{t_j}\right], \left[\frac{r_i}{w_j}\right], \left[\frac{s_i}{t_j}\right], \left[\frac{s_i}{w_j}\right], \left[\frac{t_i}{r_j}\right], \left[\frac{t_i}{s_j}\right], \left[\frac{w_i}{r_j}\right], \left[\frac{w_i}{s_j}\right] $$
Now, we use the Law of Quadratic Reciprocity: $$ \left[\frac{p}{q}\right] \left[\frac{q}{p}\right] = (-1)^{(p-1)(q-1)/4}, $$ for all the pairs involved as before. If either $p,q\equiv 1 \pmod 4$, then the product of symbols is $1$.
For the remaining case $p,q\equiv 3 \pmod 4$, the produt of symbols is $-1$. This happens between parings of $s_i$ and $w_j$, which has $b$ and $d$ elements respectively. Therefore the product of the symbols is $(-1)^{bd}$.
Equating the two different ways of computing the product of symbols, we obtain $$ (-1)^{b+d} = \prod_{p|u,q|v \text{ or } p|v, q|u} \left[\frac{p}{q}\right] = (-1)^{bd} $$ which completes the proof. $$ \tag*{$\square$} $$
Our final step is as follows:
Proof. By considering the parity of $b$ and $d$, we see that $$ (-1)^{bd} = (-1)^{b+d} $$ can hold if and only if $b$ and $d$ are both even. Noting that $r_i\equiv 1\pmod 4$ and $s_i\equiv 3\pmod 4$, this gives us $$ u = (r_1r_2\cdots r_a)(s_1s_2\cdots s_b) \equiv (1)\cdot(1) \equiv 1\pmod 4 $$ Recall that $$ k = \rho^2 u $$ Since $k$ is odd, $\rho$ is also odd and hence $\rho^2\equiv 1\pmod 4$. Therefore $ k\equiv 1 \pmod 4, $ completing the proof. $$ \tag*{$\square$} $$
We ended with a contradiction on the condition of $k$, getting $k\equiv 1\pmod 4$, therefore there cannot be an integral solution.
Other case: $2$ divides $m$ an odd number of times
Proposition 1 still holds, except that now $v$ is even. We first show that
Proof. Taking $\pmod 2$, we see that $x$ is odd. Now taking $\pmod 4$, we have $$ \begin{align} z^2v - 1 &\equiv 3 + vy^2 \pmod 4 \\ z^2v &\equiv vy^2 \pmod 4\\ z^2(v/2) &\equiv (v/2)y^2 \pmod 2\\ z &\equiv y \pmod 2 \end{align} $$ where the last line is because $v/2$ is odd (since $v$ is squarefree). If $z,y$ are both even then $$ -1 \equiv ux^2 \equiv u \pmod 8 $$ Otherwise $$ \begin{align} z^2u^2v -1 &\equiv ux^2 + vy^2 \pmod 8\\ v-1 &\equiv u+v\pmod 8\\ -1&\equiv u \pmod 8 \end{align} $$ So that for both cases, $u\equiv 7\pmod 8$. $$ \tag*{$\square$} $$
We need factorization of primes split into groups of $\pmod 8$, since their symbol $\left[\frac{2}{p}\right]$ differs:
We derive in a fashion similar to proposition 3:
Proof. As in proposition 3, the only equations that give a product as a power of $-1$ are the ones $\equiv 3\pmod 4$. These are $s_i,f_i,w_i$ with $b,h,d$ elements each, giving $b+h+d$ equations, hence the overall product $(-1)^{b+h+d}$.
For the other way of computing product of all symbols, if we ignore prime $2$ first then a similar derivation for $\equiv 3\pmod 4$ primes gives $$ (-1)^{bd}\cdot (-1)^{hd} = (-1)^{bd+hd} $$ This is simply the number of ways to pair $s_i$ with $w_j$ and $f_i$ with $w_j$. For prime $2$, the Legendre symbol $$ \left[\frac{2}{p}\right] $$ is $-1$ only when $p\equiv 3,5\pmod 8$, corresponding to primes $s_i$ and $e_i$. There are $b$ and $g$ elements respectively, giving a product of $(-1)^{b+g}$. Therefore putting everything together we have $$ (-1)^{b+h+d} = (-1)^{bd+hd+b+g} $$ $$ \tag*{$\square$} $$
The final part of this case:
Proof. We note that $$ \begin{align} u &= (\prod_{i=1}^a r_i)(\prod_{i=1}^bs_i)(\prod_{i=1}^ge_i)(\prod_{i=1}^hf_i)\\ u &\equiv 3^b5^g7^h \pmod 8 \end{align} $$ Equating the indices of proposition 7: $$ \begin{align} b+h+d &\equiv bd+hd+b+g \pmod 2\\ d(b+h+1) &\equiv h+g \pmod 2 \end{align} $$ If $d\equiv 0\pmod 2$, then $h\equiv g\pmod 2$. If $h,g$ even: $$ u \equiv 3^b \not\equiv 7\pmod 8 $$ else $h,g$ odd: $$ u\equiv 3^b \cdot 5\cdot 7 \equiv 3^{b+1}\not\equiv 7\pmod 8 $$
Alternatively, $d\equiv 1\pmod 2$ and $$ b+h+1 \equiv h+g \pmod 2 \implies b+1 \equiv g \pmod 2 $$ If $b$ odd and $g$ even: $$ u \equiv 3\cdot 1\cdot 7^h \equiv 3,5 \not\equiv 7 \pmod 8 $$ otherwise $b$ even and $g$ odd: $$ u \equiv 1\cdot 5 \cdot 7^h \equiv 3,5 \not\equiv 7\pmod 8 $$
Therefore in all cases $u\not\equiv 7 \pmod 8$.
$$ \tag*{$\square$} $$
This contradicts the initial assumption of $u\equiv 7\pmod 8$, therefore there cannot be an integral solution.