Define funcitons F, G and H:
F(x,y) = x
^{2}  2xy + y
^{2}  5x + 7y
G(x,y) = x
^{2}  3xy + 2y
^{2} + x  y
H(x,y) = xy  12x + 15y
If x and y are integers and p is an odd prime such that the values of F(x,y) and G(x,y) are both divisible by p, then prove that H(x,y) is also divisible by p.
Seventeen divides expression? is a specific instance of this puzzle where p=17.
Broll has solved it full, anyway it's worth also the direct approach.
As broll says G=(xy)(x2y+1). To have G multiple of p, one of both factors should be multiple of p.
a) (xy) multiple of p.
F=(xy)^25x+7y. Then to have F also divisible by p for values (xy)=0 (mod p) it should be that 7y5x =0 (mod p). From both in bold, 2y is 0 (mod p) (=> y is 0 mod p) and x is 0 (mod p).
=>(x,y) are both 0 (mod p). H is divisible by p in this case.
b) (x2y+1) multiple of p
Playing a little, see that:
F=G + H/2 +y(x2y+1)/2
To have F also divisible by p (for values (x2y+1) = 0 (mod p)) is necessary that also H is 0 (mod p) for that values.
 This is because considering the three summands: G is multiple of p and y(x2y+1)/2 is a multiple of p or at least half a multiple of p. Only if also H is multiple of p, F will also be.
 Note also that F is always and integer (for x, y integers)
Conclusion: On the basis of G divisible by p, to have F divisible by p implies necessarily H divisible by p. As it is a given data that F is multiple of p for certain x,y values, the same values for which G is divisible by p, then, for that values, necessarily H should be divisible by p.
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The seventeen divides expression puzzle is not fully solved as broll's solution considers a wide set of answers.
I can add that if p=17 and I restricted (x2y+1) to be not only a p multiple, but just p, I mean (x2y+1)=p*1=17 it's easy to obtain the solution x=20 y=2, not considered in the above mentioned puzzle.
F(20,2)=238=17*14
G(20,2)=306=17*18
H(20,2)=170=17*10
Edited on July 22, 2016, 11:17 am

Posted by armando
on 20160722 04:10:53 