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.
I never set out the full proof of the last problem, Seventeen divides expression?
1. x^22xy+y^25x+7y = (xy)^25x+7y
2. x^23xy+2y^2+xy = (xy)(x2y+1)
From the factorisation of 2, there are 2 cases:
xy=17k
x2y+1 =17k
Since 17 is prime, it must appear wholly in one of the two factors.
First case: (xy)=17k
Substituting in 1: (17k)^25x+7y, so 17 divides 7y5x
7y5x+5(xy) = 2y Adding multiples of (xy)
So 2y is a multiple of 17; but if so then y is a multiple of 17 and so is x:
7y5x+7(xy) =2x Adding multiples of (xy)
Therefore 17 divides xy12x+15y
This part works for any odd prime.
Second case: similar method, but a bit more difficult.
If (x2y+1) =17k, then (xy) = 17k+y1
(x^22xy+y^25x+7y)+5(x2y+1) = (xy)^23y+5
Adding multiples of (x2y+1)
Substituting for 17k+y1 for (xy):
289k^2+34ky34k+y^25y+6 = 17k(17k+2y2)+y^25y+6
Then y^25y+6 =17n, so y=17m+2, 17m+3
if y=17m+2, then x=17k+34m+3 = 17n+3
If y=17m+3, then x=17k+34m+5 =17n+5
Checking: (17n+32(17m+2)+1) = 17(n2m), which is divisible by 17.
Either xy12x+15y = (17m+2)(17n+3)12(17n+3)+15(17m+2) = 17(17mn+18m10n), or
xy12x+15y = (17m+3)(17n+5)12(17n+5)+15(17m+3) = 17(17mn+18m7n)
In either case, 17 still divides xy12x+15y.
Again, under substitution, this part works for any odd prime.

Posted by broll
on 20160721 00:05:20 