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:     
x-y=17k
x-2y+1 =17k 

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