<s>[INST] 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.
 [/INST] 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 $6$.
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