Question

Q: 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.

Answer 1

Anyone?

Answer 2

since KL+MN=(K+M-L+N)(-K+M+L+N), then we know that will not only be divisible by one and itself. It will also be divisible by K+M-L+N and -K+M+L+N since it is also a product of these too.

Answer 3

How did you get KL+MN=(K+M-L+N)(-K+M+L+N)?

Answer 4

No
\[(K+M-L+N) (-K+L+M+N)=\\-K^2+2 K
L-L^2+M^2+2 M N+N^2
\]

Answer 5

@eliassaab what do you mean?

Answer 6

I mean
\[ KL+MN\ne(K+M-L+N)(-K+M+L+N)
\]

Answer 7

because KM+LN=(K+L-M+N)(-K+M+N+L) so I've mistaken that for
KL+MN=(K+M-L+N)(-K+M+L+N)

Answer 8

sorry :))

Answer 9

since we know KM + LN is not prime, if we can show it shares a common factor with
KL + MN i think that will do it

Answer 10

@eigenschmeigen how do I do it?

Answer 11

im not sure, i havent really attempted anything

Answer 12

Please help!

Answer 13

@nikvist

Answer 14

i stuck the first equation in wolfram, says there aren't any positive integer solutions
what does this mean? are we still able to prove this?

http://www.wolframalpha.com/input/?i=ab+%2B+cd+%3D+%28a+%2B+c+-+b+%2B+d%29%28-a+%2B+c+%2B+b++%2B+d%29

Answer 15

@aron_west

Answer 16

Yes, finally someone is replying!

Answer 17

@eliassaab , in expanding the right side of "KM + LN = (K + L - M + N)(-K + L + M + N)", i got 2KM + 2LN terms along with some other terms....

anybody got that too?

Answer 18

Me, Me ...MEEEHH! I got that as well.
\[KM + LN = K^2 + M^2 - L^2 - N^2\]

Answer 19

yah, thats what i got...

Answer 20

http://www.wolframalpha.com/input/?i=+expand+%28K+%2B+L+-+M+%2B+N%29+%28-K+%2B+L+%2B+M+%2B+N%29

Answer 21

I posted this problem on m.se, none got the solution there. I think something is not right.

Answer 22

@eigenschmeigen That means the problem has something not right with it.

Answer 23

Nothing is wrong with the problem, it's an IMO problem or Non-solvable by commons like me.

http://imo.wolfram.com/problemset/IMO2001_solution6.html