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Suppose we want to solve $4(x+y)^{2}-3xy-6(x+y)=0$ where $x$ and $y$ are both integers. Why we only get the trivial solution?

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The equation can be arranged as a quadratic equation of x: $4x^2+x(5y-6)+(4y^2-6y)=0$.

The discriminant(D)= $(5y-6)^2-4.4.(4y^2-6y)=36+36y-29y^2$

Now D needs to be ≥0 as x is real.

$36+36y-29y^2 ≥0 ⇔ 29y^2-36y-36 ≤0 $

=>$\frac{18-6\sqrt{38}}{29}≤y≤\frac{18+6\sqrt{38}}{29}$

as $(t-a)(t-b)≤0$(where a≤b) => (a≤t and t≤b) or (b≤t and t≤a).

Clearly, a≤t≤b as the other option is impossible.

Now, $\frac{18-6\sqrt{38}}{29}>-1$ and $\frac{18+6\sqrt{38}}{29}<2$ (by observation)

$=>-1 < y < 2$ as y being integer can be 0 or 1 to make x real.

If $y=0,D=36$, if $y=1,D=43$ .

Also D needs to be perfect square as x is integer.

So, y must be 0 to make x real ad integral.

Observation: the function in the LHS of the equation is symmetric w.r.t. x,y. The calculation & conclusion won't change if we interchange x,y.

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  • $\begingroup$ Also $D < 0$ if $y \le -1$. $\endgroup$ – Robert Israel Aug 5 '12 at 7:25
  • $\begingroup$ @RobertIsrael, thanks for your observation, I have included it $\endgroup$ – lab bhattacharjee Aug 5 '12 at 7:28
  • $\begingroup$ Can anybody please explain me what is this 'community wiki'. $\endgroup$ – lab bhattacharjee Aug 6 '12 at 4:59
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We use one of my favourite identities, $4xy=(x+y)^2-(x-y)^2$. Let $x+y=s$ and $x-y=t$.

To avoid fractions, multiply our equation through by $4$. We get $$16(x+y)^2-12xy-24(x+y)=0,$$ which can be rewritten as $$13s^2 +3t^2-24s=0.\tag{$1$}$$ For Equation $(1)$ to hold, we need $13s^2-24s \le 0$. The real zeros of $13s^2-24s$ are at $s=0$ and $s=24/13$, so the only possible integer values of $s$ are $s=0$ and $s=1$.

If $s=0$, then for $(1)$ to hold we need $t=0$, giving $x=y=0$. If $s=1$, we need $3t^2=11$, which is impossible.

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  • $\begingroup$ another contradiction: $3|3t^2 \implies 3|(13s^2-24s) \implies 3^2|13s^2-24s \implies 3|t \implies 3|\gcd(s,t)$ $\endgroup$ – miracle173 Mar 3 '16 at 4:48

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