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If $M,N,P$ are positive integers such that $$\begin{cases}M+N+P&=2024\\MNP&=2020^2\end{cases},$$

Show that $(M,N,P)=(4,1010,1010)$ is the only solution up to permutation.

I've got this problem from another problem, and I prove this by using brute-force. I assume $M\leq N\leq P$ to deduce that $M\leq \sqrt[3]{2020^2}<160$, so the divisors of $2020^2$ bounded by $160$ are $$M=1,2,4,5,8,10,16,20,25,40,50,80,100,101.$$ Then, I put each $M$ to obtain $$\begin{cases}N+P=2024-M\\NP=2020^2/M\end{cases}$$ which $N,P$ are the two roots of a quadratic equation, and $N,P$ will be positive integers only if the discriminant of the quadratic equation is a perfect square. At this step I've used programming to verify that, besides $M=4$, other discriminants are not perfect squares.

I'm wondering is there any mathematical way to prove this. More generally, it will be natural to conjecture that if $K\geq 1$ is a positive integer, then the only positive integer solutions $M,N,P$ to $$\begin{cases}M+N+P&=2K+4\\MNP&=4K^2\end{cases}$$ are $(M,N,P)=(4,K,K)$ up to permutation. A proof for this conjecture will also be appreciated.

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$M, N, P$ must contain a total of two factors of $101$ in their prime factorization, and they must be less than $101^2$ to satisfy the sum equation. Therefore two of them must he multiples of $101$. Wlog call those two $N, P$. Then the sum equation forces $M\equiv 4\bmod 101$ and $M$ has to be a divisor of $2020^2$. Only $M=4$ meets both requirements. A number that is not a multiple of $101$ and divides $2020^2$ must divide $20^2=400$, and out of $\{4,105,206,307\}$, only $4$ does so.

Once $M=4$ is proved, $N$ and $P$ are easily shown to be the roots of $x^2-2020x+1010^2=0$ forcing both to be $1010$.

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    $\begingroup$ Indeed elegant. Thanks! By the way, do you have any way to solve the general case? $\endgroup$
    – kelvin hong 方
    Commented Mar 14, 2020 at 12:38
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    $\begingroup$ Following your argument, $N,P$ are multiples of $101$ it no longer is immediate that $M$ is the smallest of the three and therefore the checking that $M$ is the only $4 \mod 101$ divisor of $2020^2$ may require brute force as well. $\endgroup$
    – enochk.
    Commented Mar 15, 2020 at 2:59
  • $\begingroup$ @enochk it seems like the case, I'll investigate more cases on this problem. $\endgroup$
    – kelvin hong 方
    Commented Mar 15, 2020 at 8:38
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    $\begingroup$ But I do not assume $M\le N\le P$. $\endgroup$
    – Oscar Lanzi
    Commented Mar 15, 2020 at 8:59
  • $\begingroup$ The question is whether one can bypass the need to check all those values of $M$ by finding a faster way to prove that $M=4$. If you don't assume $M$ is the least of the three, then checking that $M \equiv 4 \mod 101$ and $M | 2020^2$ won't be faster. $\endgroup$
    – enochk.
    Commented Mar 15, 2020 at 9:15

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