Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
    
Also $D < 0$ if $y \le -1$. –  Robert Israel Aug 5 '12 at 7:25
    
@RobertIsrael, thanks for your observation, I have included it –  lab bhattacharjee Aug 5 '12 at 7:28
    
Can anybody please explain me what is this 'community wiki'. –  lab bhattacharjee Aug 6 '12 at 4:59
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.