Prove that $n$ is Divisible by $6$ when the Quadratics have integer roots If $m,n \in \mathbb{Z}$ and Both the Quadratics $x^2+mx-n=0$ and $x^2-mx+n=0$ has integer roots, Prove that $n$ is a Multiple of $6$
My try:
Let $n$ is not a multiple of $6$
We have
$$n=6q\pm1$$
OR
$$n=6q\pm 2$$
OR
$$n=6q\pm 3$$
Now since the Discriminants of both the quadratics should be a perfect square we have
$$m^2+4n=p^2$$
$$m^2-4n=r^2$$
So we get
$$8n=(p-r)(p+r)$$
$\implies$
$$6n+2n=(p-r)(p+r)$$
$$6n+2(6q\pm s)=(p-r)(p+r)$$
$$6a\pm2s=(p-r)(p+r)$$
where $s=1,2,3$
Any clue from here?
I already know this Question has an existing thread, but i will be happy if i can continue with my solution.

You mention in your question that it already "has an existing thread". You should specify where this is to help avoid people duplicating efforts. However, if it's Prove that $n$ is divisible by $6$, note my solution below is somewhat different and, as you asked, I try to continue your solution.
You started with the two discriminant equations of
$$m^2+4n=p^2 \tag{1}\label{eq1}$$
$$m^2-4n=r^2 \tag{2}\label{eq2}$$
and then proceeded to take the difference of them to get
$$6a\pm2s=(p-r)(p+r) = p^2 - r^2 \tag{3}\label{eq3}$$
where $n = 6q \pm s$ and $s=1,2,3$.
One way to proceed would be to check each case to show it results in a situation which is not possible. First, start with adding $s = 1$. In case you're not familiar with it, basic modulo notation of $x \equiv y \pmod z$ means that $x - y$ is an integral multiple of $z$. For all integers $x$, $x^2 \equiv 0 \text{ or } 1 \pmod 3$. The LHS (left hand side) of \eqref{eq3} gives $6a + 2 \equiv 2 \pmod 3$. On the right side, only $p^2 \equiv 0 \pmod 3$ and $r^2 \equiv 1 \pmod 3$ matches. This causes \eqref{eq1} to become $m^2 + 1 \equiv p^2 \equiv 0 \pmod 3$, so $m^2 \equiv 2 \pmod 3$ which is not possible.
Next, with adding $s = 2$, the LHS of \eqref{eq3} gives $6a + 4 \equiv 1 \pmod 3$. On the right side, only $p^2 \equiv 1 \pmod 3$ and $r^2 \equiv 0 \pmod 3$ matches. This causes \eqref{eq2} to become $m^2 - 8 \equiv m^2 + 1 \equiv r^2 \equiv 0 \pmod 3$, so $m^2 \equiv 2 \pmod 3$ which is not possible.
You can similarly show that subtracting $s = 1, 2$ also doesn't work. Finally, consider the case where $s = 3$, which means $n = 6q + 3$ is odd. For this, I don't see any direct way to use your final equation since it's lost some important information. Instead, from \eqref{eq1}, moving $m^2$ to the other side gives
$$4n = p^2 - m^2 \tag{4}\label{eq4}$$
The LHS side is even so both $p$ and $m$ must be either odd or even. For the first case, note for all odd integers $x$ that $x^2 \equiv 1 \pmod 8$. As such, $p^2 - m^2 \equiv 0 \pmod 8$. However, $4n \equiv 0 \pmod 8$ if $n$ is even and $4n \equiv 4 \pmod 8$ if $n$ is odd, so the values only match when $n$ is even. For the case where $p$ and $m$ are both even, then consider \eqref{eq1} plus \eqref{eq2}:
$$2m^2 = p^2 + r^2 \tag{5}\label{eq5}$$
Since $m$ and $p$ are even, so is $r$. Thus, if $p = 2p_1$, $m = 2m_1$ and $r = 2r_1$, you can divide by a factor of $4$ to get
$$2m_1^2 = p_1^2 + r_1^2 \tag{6}\label{eq6}$$
Now, both $p_1$ and $r_1$ must be both either even or odd. If they're even, then $p_1^2 + r_1^2$ has a factor of $4$, so $m_1$ must be even. In this case, $p$ and $m$ must each be a multiple of $4$ so $p^2 - m^2$ is a multiple of $16$ meaning $n$ is even in \eqref{eq4}. If both $p_1$ and $r_1$ are odd, then their squares are each congruent to $1$ modulo $4$, so $m_1$ must be odd. Thus, in \eqref{eq4}, $p^2 - m^2 = 4(p_1 + m_1)(p_1 - m_1)$ must be a multiple of $16$. As such, in either case, $n$ must be even, so $s = 3$ doesn't work.
This shows that $s$ must be $0$, i.e., $n$ is a multiple of $6$. However, a somewhat simpler & more direct way to have accomplished this is to use the method above to show that $n$ must be even and then check modulo $3$. In particular, if $n \equiv 1 \pmod 3$, \eqref{eq1} gives that $m^2 + 1 \equiv p^2 \pmod 3$. Since $m^2 \equiv 1 \pmod 3$ doesn't work as this gives that $p^2 \equiv 2 \pmod 3$, then $m^2 \equiv 0 \pmod 3$. However, \eqref{eq2} then gives that $0 - 1 \equiv 2 \equiv r^2 \pmod 3$, which can't be true. Similarly, if $n \equiv 2 \pmod 3$, then $m^2 \equiv 1$ from \eqref{eq1}, but this again requires that $r^2 \equiv 2 \pmod 3$ from \eqref{eq2}. Thus, the only choice left is that $n \equiv 0 \pmod 3$. Since $n$ must have a factor of $2$ and a factor of $3$, it's a multiple of $6$.
