If(x,y)integers&?

Question

if f(x,y)integers&17divides both-xsquare-2xy+ysquare-5x+7y,xsquare-3xy+2ysquare+x-y then show xy-12x+15yis also divisible by 17

Answer 1

Ever hear of a spacebar? And you don't write "square" to indicate squaring, you write ² or ^2.

Anyway, onto the problem:

x² - 2xy + y² - 5x + 7y ≡ 0 mod 17
x² - (2y+5)x + y² + 7y ≡ 0 mod 17
x² - (2y+5)x ≡ -y² - 7y mod 17
Now we complete the square. First we find (2y+5)/2 mod 17, which is y+11 (since 11*2 ≡ 5 mod 17). Now, we square it, to yield y²+5y+2 (again, mod 17).
x² - (2y+5)x + y² + 5y + 2 ≡ -y² - 7y + y² + 5y + 2 mod 17
(x-y+11)² ≡ -2y + 2 mod 17
x ≡ y - 11 ± √(-2y+2) mod 17

Now, we substitute this into the other congruence:

x² - 3xy + 2y² + x - y ≡ 0 mod 17
(y - 11 ± √(-2y+2))² - 3(y - 11 ± √(-2y+2))y + 2y² + y - 11 ± √(-2y+2) - y ≡ 0 mod 17

Multiplying everything out:

y² - 7y ± 2y√(-2y+2) + 4 ± 12√(-2y+2) - 3y² + 33y ± 14y√(-2y+2) + 2y² + y - 11 ± √(-2y+2) - y ≡ 0 mod 17

And simplifying:

9y - 7 ± (16y+13)√(-2y+2) ≡ 0 mod 17
± (16y+13)√(-2y+2) ≡ 7-9y mod 17

Squaring both sides:

(y² + 8y + 16)(-2y+2) ≡ 13y² - 7y + 15 mod 17
-2y³ - 16y² - 15y + 2y² + 16y + 15 ≡ 13y² - 7y + 15 mod 17
-2y³ - 14y² + y + 15 ≡ 13y² - 7y + 15 mod 17
-2y³ - 10y² + 8y ≡ 0 mod 17
-y(2y² + 10y - 8) ≡ 0 mod 17
Since 17 has no zero divisors, this implies that y≡0 mod 17 or 2y² + 10y - 8 ≡ 0 mod 17. In the latter case:

2y² + 10y - 8 ≡ 0 mod 17
y² + 5y - 4 ≡ 0 mod 17
y²+5y+2 ≡ 6 mod 17
(y+11)² ≡ 6 mod 17

However, 6 has no square roots mod 17, so this is impossible. So y≡0 mod 17. We may now go back and solve for x:

x ≡ y - 11 ± √(-2y+2) mod 17
x ≡ - 11 ± √2 mod 17
x ≡ -11 ± 6 mod 17
x ≡ 6 ± 6 mod 17
x ≡ 0 mod 17 or x≡12 mod 17

We have two possible solutions - it is best that we test them on your original congruences, which are:

x² - 2xy + y² - 5x + 7y ≡ 0
x² - 3xy + 2y² + x - y ≡ 0

Clearly, these are both satisfied by the solution x≡0, y≡0. However, we have a second possible solution, which is x≡12, y≡0. Testing it on the first congruence gives us:

x² - 2xy + y² - 5x + 7y ≡ 0
8 - 9 ≡ 0 mod 17

Which is clearly not true. So the only way the given congruence relations can hold is if x≡0 and y≡0. Substituting this solution into xy-12x+15y trivially yields that this expression is congruent to zero, and we are done.