Define funcitons F, G and H:
F(x,y) = x2
- 2xy + y2
- 5x + 7y
G(x,y) = x2
- 3xy + 2y2
+ 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^2-2xy+y^2-5x+7y = (x-y)^2-5x+7y
2. x^2-3xy+2y^2+x-y = (x-y)(x-2y+1)
From the factorisation of 2, there are 2 cases:
Since 17 is prime, it must appear wholly in one of the two factors.
First case: (x-y)=17k
Substituting in 1: (17k)^2-5x+7y, so 17 divides 7y-5x
7y-5x+5(x-y) = 2y Adding multiples of (x-y)
So 2y is a multiple of 17; but if so then y is a multiple of 17 and so is x:
7y-5x+7(x-y) =2x Adding multiples of (x-y)
Therefore 17 divides xy-12x+15y
This part works for any odd prime.
Second case: similar method, but a bit more difficult.
If (x-2y+1) =17k, then (x-y) = 17k+y-1
(x^2-2xy+y^2-5x+7y)+5(x-2y+1) = (x-y)^2-3y+5
Adding multiples of (x-2y+1)
Substituting for 17k+y-1 for (x-y):
289k^2+34ky-34k+y^2-5y+6 = 17k(17k+2y-2)+y^2-5y+6
Then y^2-5y+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+3-2(17m+2)+1) = 17(n-2m), which is divisible by 17.
Either xy-12x+15y = (17m+2)(17n+3)-12(17n+3)+15(17m+2) = 17(17mn+18m-10n), or
xy-12x+15y = (17m+3)(17n+5)-12(17n+5)+15(17m+3) = 17(17mn+18m-7n)
In either case, 17 still divides xy-12x+15y.
Again, under substitution, this part works for any odd prime.
Posted by broll
on 2016-07-21 00:05:20