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Composite Contemplation (Posted on 2013-02-16) Difficulty: 3 of 5
S=a+b+c+d+e+f, where:
each of a,b,c,d,e and f is a positive integer.

If S divides each of:
a*b*c + d*e*f and:
a*b + b*c + c*a - d*e - e*f - d*f
then, is S always composite?

If so, prove it. If not, provide a counter example.

No Solution Yet Submitted by K Sengupta    
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spoiler | Comment 1 of 3
Assume S is prime and derive a contradiction.

Add the 3 congruences abc+def = ab+bc+ac-de-ef-df = a+b+c+d+e+f = 0 mod S to get the factoring (a+1)(b+1)(c+1) + (d-1)(e-1)(f-1) = 0 mod S.

This process is repeatable, so we get (a+k)(b+k)(c+k) + (d-k)(e-k)(f-k) = 0 mod S for positive k.

At some point one of the factors of the second term becomes zero.  

Let's say k=d, giving (a+d)(b+d)(c+d) = 0 mod S.

Since S is prime at least one of these factors is a multiple of S. 

Say a+d = mS, m>=1.

But then the condition that a+b+c+d+e+f = S requires b+c+e+f = 0, when they're all supposed to be positive.  There's the contradiction.

  Posted by xdog on 2013-02-17 15:35:12
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