Solve the equation in integers numbers $3x^2-2xy-7=y^2$.

Question

PROBLEM:

Solve the equation in integers numbers $3x^2-2xy-7=y^2$.

WHAT I THOUGHT OF:

What if we pass $y^2$ to the other member of the equality.

$3x^2-2xy-7-y^2=0$

Forward it would be simple if we can write $3x^2-2xy-7-y^2$ as a product of 2 equations. I need help here, because i thought a lot about how can i rewrite it. Also, abbreviated calculation formulas might help. Hope one of you can help me!

Answer 1

Observe that by moving $y^2$ to the $LHS$, we can factorise as $(x - y)(3x + y) = 7$. I typically spot these kinds of factorisations by considering the case $y = 1$; alternatively, you can use the fact that it can be rewritten as $(x - y)^2 + 2(x^2 - y^2)$, from which the factorisation is more readily apparent.

Since $x$, $y$ are integers, it follows these brackets are integers. Hence, we must have either \begin{align*} x - y = -7&, 3x + y = -1 &&\implies (x, y) = (-2, 5)\\ x - y = -1&, 3x + y = -7 &&\implies (x, y) = (-2, -1)\\ x - y = 1&, 3x + y = 7 &&\implies (x, y) = (2, 1)\\ x - y = 7&, 3x + y = 1 &&\implies (x, y) = (2, -5) \end{align*} Hence, all solutions are $(-2, 5), (-2, -1), (2, 1), (2, -5)$.

Answer 2

First, $3x^2−2xy−7−y^2$ isn't an equation: that's an expression. $3x^2−2xy−7−y^2 = 0$ is an equation.

Now let's say you want to write $3x^2−2xy−7−y^2$ as a product of two expressions. You can start by looking for any solutions, integer or not, complex or not.

Here, notice that if you fix y you have a nice 2nd degree polynomial in x: $ax^2+bx+c=0$ (with $a=3$, $b=-2y$, $c=-7-y^2$).

Just apply the formula for discriminant: $\Delta = b^2-4ac = 4y^2+12(7+y^2) = 16y^2+84$

$\Delta > 0$, so $\frac {-b+\sqrt{\Delta} }{2a}$ and $\frac {-b-\sqrt{\Delta} }{2a}$ are roots: $3x^2−2xy−7−y^2 = (x-\frac {-b+\sqrt{\Delta} }{2a})(x-\frac {-b-\sqrt{\Delta} }{2a}) = (x-\frac {y+\sqrt{4y^2+21} }{3})(x-\frac {y-\sqrt{4y^2+21} }{3})$

And thus, the equation $3x^2−2xy−7−y^2 = 0$ is equivalent to $(x-\frac {y+\sqrt{4y^2+21} }{3})(x-\frac {y-\sqrt{4y^2+21} }{3}) = 0$

You could have done the same thing by fixing x instead of y, and that would have given you a different result (but the same set of solutions).

Answer 3

$3x^{2}-2xy-7=y^{2}$ then $(\sqrt{3} x)^{2} +(\frac{1}{\sqrt{3}} y)^{2} -(1+\frac{1}{3})y^{2}-2xy-7=0$

i.e $7=(\sqrt{3} x -\sqrt{\frac{1}{3}} y -\frac{2}{\sqrt{3}} y)(\sqrt{3} x -\sqrt{\frac{1}{3}} y +\frac{2}{\sqrt{3}} y)=(x-y)(3x+y)$ then , 1).if $x-y=1$ then $3x+y=7$ conclution solution is $(x,y)=(2,1)$ and if $x-y=7$ then $3x+y=1$ then solution,$(x,y)=(2,-5)$

$x-y=-1$ then $3x+y=-7$ Solution:$(x,y)=(-2,-1)$

$x-y=-7$ then $3x+y=-1$ Solution:$(x,y)=(-2,5)$

then solution general is $(x,y)=\{(-2,-1),(-2,5),(2,1),(2,-5)\}$.