# Number Theory Problem.

K $>$ L $>$ M $>$ N are positive integers such that$,$ KM $+$ LN $=$ (K$+$L$-$M$+$N)($-$K$+$L$+$M$+$N)$.$ Prove that KL $+$ MN is not prime.

I'm stumped :/

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@Chandrasekhar (assuming K+L-M+N, -K+L+M+N > 1) –  Ronald Apr 15 '12 at 10:49
@Chandrasekhar Are you sure, you read the question correctly? Because I don't see how $KL+MN$ is divisible by $K+L-M+N$. –  Ishaan Singh Apr 15 '12 at 10:57
@IshaanSingh: Sorry –  user9413 Apr 15 '12 at 11:00
@Ronald $KM + LN$ is divisible by $K+L-M+N$, not $KL+MN$. –  Ishaan Singh Apr 15 '12 at 11:15
@IshaanSingh oh! apologies! a nice question, then. –  Ronald Apr 15 '12 at 12:39

This is problem 6 from the 2001 Intnernational Mathematics Olympiad. That should be enough for you to google the answer (or perhaps shame you into not doing so).

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$L-M$ is positive so $K+L-M+N$ is positive, and hence the second factor $-K + L + M +N$ is also positive. Now assume $KL + MN$ is prime. Then one of the factors is $1$ and the other is $KL + MN.$ Since $K+L - M + N > K+N > 1$ we have $$K + L - M + N = KL + MN.$$

This rearranges to $L-M = K(L-1) + (M-1)N$ which is absurd since the left side is less than $L$ while the right hand side is at the very least $K > L.$ Thus the assumption leads to a contradiction and the number is prime.

EDIT: Now that I realize this is a Problem 6 from an IMO competition, I suspect I've made a grave error in my answer above, but I can not see it. I would be grateful if someone could point it out.

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Question is about $KL+MN$, and lhs of given equality is $KM+LN$ –  Julius Apr 15 '12 at 13:17
@Julius Ahh yup, I'm silly! Thank you. –  Ragib Zaman Apr 15 '12 at 13:19
The second solution in the link KV Raman posted rearranges the condition to $a^2-ac + c^2 = b^2+bd + d^2$ before proceeding. Those are the norms of $a+c \omega$ and $b-d\omega$ in the Eisenstein integers $\mathbb{Z}[\omega].$ I wonder if there is an illuminating solution using properties of the Eisenstein integers? –  Ragib Zaman Apr 15 '12 at 13:40