# Finding positive integer solutions to $n = ax^2 +by^2 - cxy$

How can I find the positive integer solutions to $x$ and $y$, given that $n$, $a$, $b$ and $c$ are all positive integers, in an equation of the form:

$$n = ax^2 + by^2 - cxy.$$

Specifically, I want to find the positive integer solutions to the following equation, given $n$:

$$n=3 x^2+20 y^2-16 x y.$$

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$$n = (3x - 10y)(x-2y) {}{}$$

Alright, changing variables with $$u = x - 4 y, \; v = x - 3 y$$ so that $$x = 4 v - 3 u, \; y = v - u,$$ we have $$3 x^2 - 16 x y + 20 y^2 = -u^2 + 4 v^2.$$

So, factoring $n,$ we find all possible ways to write $$n = -u^2 + 4 v^2 = (2 v + u )(2 v - u)$$ as an even square minus an odd square, definitely including both $u,-u$ for each success, also both $v,-v.$ Then, for each success (finitely many) we switch back to the original variables with $x = 4 v - 3 u, \; y = v - u,$ and choose the solutions that include your conditions on positivity, whatever you might have meant by that.

Note that there are no solutions if $n \equiv 1,2 \pmod 4.$ When $n<0,$ there are no solutions when $|n| \equiv 2,3 \pmod 4.$

EDIT, Wednesday morning, before Gerry wakes up in Australia, another completely cosmetic change: switch to $-n$ and find all solutions to $$-n = s \, (s + 4 t)$$ which is just to find ALL ways of writing $$-n = FG$$ such that $F$ and $G$ differ by a multiple of $4.$ Then use $$x = -3s - 2 t, \; y = -s-t,$$ from $$s = -x+2y, \; t = x-3y.$$ So There.

Well, if $n=0$ I guess there really are infinitely many solutions. But, as soon as $n \neq 0,$ we get finitely many solutions because $$|s| \leq |n|, \; \; |s+4t| \leq |n|.$$

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How will this find me the integer solutions to x and y? –  Khaled Aug 7 '12 at 20:13
@Khaled: Hint: factor $n$ –  robjohn Aug 7 '12 at 20:19
So if $n = ab$ then $x = (5b-a)/2$ ? –  Khaled Aug 7 '12 at 20:26
@Khaled, no. I just typed in a very full solution. Try three examples, $n=7,\; 11, \; 15$ by hand. I see, you said you wanted $\langle 3,-16,20\rangle$ so that is what I did. There are no $a,b,c$ in my solution. –  Will Jagy Aug 7 '12 at 20:39
Will, I don't see what's gained in this example by changing variables. For every factorization $n=rs$, you get the system $3x-10y=r$, $x-2y-s$, and you see whether it has an integer solution. –  Gerry Myerson Aug 8 '12 at 1:31
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$ax^2+by^2-cxy=n$

Expressing as a quadratic equation of x, $ax^2-x(cy)+by^2-n=0$

As x is positive integer, $=>(cy)^2-4.a(by^2-n)=(c^2-4ab)y^2 +4an$ must be perfect square.

$c=16, a=3, b=20 =>16y^2+12n=d^2(say)$=>d is even=2e(say)

$=>4y^2+3n=e^2$=>e is odd $>e^2≡1(mod\ 8)$

$=>e^2-4y^2 \equiv 1 \pmod4$ => $3n \equiv 1 \pmod 4$=>$n\equiv -1 \pmod 4$ to admit solution.

So, n=4m-1 for some integer m.

Applying the the approach on y, $(c^2-4ab)x^2 +4bn$ must be perfect square,

or, $16x^2+80n=f^2$=>f must even=2g(say),

$=>4x^2+20n=g^2$=>g is even=2h(say),

$=>x^2+5n=h^2$

If h=5s for integer s, $x^2≡0(mod\ 5)$=>x=5t for some integer t.

If h=5s±1 for integer s, $x^2≡1(mod\ 5)$=>x=5t±1 for some integer t.

If h=5s±2 for integer s, $x^2≡4(mod\ 5)$=>x=5t±2 for some integer t.

n will be $\frac{h^2-x^2}{5}$

$=>x^2+5(4m-1)=h^2$

If x,h are both odd or both even, $5(4m-1)=h^2-x^2≡0(mod\ 4)$ which is impossible.

If x is even, $h^2=x^2+5(4m-1)≡-1(mod\ 4)$ which is impossible.

=> x must be odd and h must be even to admit solution.

y can also calculated from the given once x is known.

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When TeX-ing congruences, you can use \pmod, which gives nicer spacing. E.g., $x\equiv1\pmod2$ produces $x\equiv1\pmod2$ and $n\equiv 0\pmod{pq}$ produces $n\equiv 0\pmod{pq}$. –  Martin Sleziak Aug 8 '12 at 6:31