A system of equations over the positive integers

Question

Find all triplets of positive integers $(k,l,m)$ such that $k,l,m \geq 2$, $k+l+m=2002$ and $k^2+l^2+m^2-klm=4$.

I observed that if the distance between the numbers is very big, for example if the distance between k and l is very big (and by distance I mean |k-l|) then $k^2+l^2+m^2-klm$ is also very big, so it can’t be 4. But I don’t know exactly if this is true. Please help me solve this system. Thank you in advance!

Answer 1

Modulo $m+2$ we have $$k+l\equiv2004,(k+l)^2\equiv0.$$

Therefore $m+2$ is a factor of $2004^2=2^4.3^2.167^2.$ The same, of course, is true for $k+2$ and $l+2$.

Since $(k+2)+(l+2)+(m+2)=2008$, we have $$\{k+2,l+2,m+2\}=\{167a,b,c\}\text { or }\{167a,167b,c\},$$ where $a,b$ and $c$ are factors of $2^4.3^2$ and are therefore no greater than $144$.

In the first case, considerations of size give $a=11$ which is not a factor of $2^4.3^2$ or $a=12,b=c=2$ when two of $k,l,m$ would be zero.

In the second case, we similarly obtain $a+b=12,c=4$. Then one of $k,l,m$, say $m$, is $2$. Then the second given equation becomes $(k-l)^2=0$ and we have $$\{k,l,m\}=\{1000,1000,2\}.$$

Answer 2

$k^2+l^2+m^2-klm=4\\ \overset{l\to2002-k-m}{\implies} (2 + k) (2 m + k - 2002)^2 = (k - 2) (k^2 - 4008 k + 4007996)\\ \overset{at\ least}{\implies} (k,m)=(2,1000)$

By symmetry $(k,l,m)=(2,1000,1000),(1000,2,1000),(1000,1000,2)$

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$(2 + k) (2 m + k - 2002)^2 = (k - 2) (k^2 - 4008 k + 4007996)\\ \implies \biggl((2 + k)(2 m + k - 2002)\biggr)^2 =k^4 - 4008 k^3 + 4007992 k^2 + 16032 k-16031984$

magma code:

IntegralQuarticPoints([1, -4008, 4007992, 16032, -16031984]);

Output:

[
    [ 1000, -999996 ],
    [ 2, 0 ]
]

I.e. no other solutions.