How to solve this equation $\ x^2-y^2=29$

Question

How to find the solutions $\ (x,y)$ in this equations

1.$\ x^2-y^2=29$

2.$\ x^2+xy+y^2=0$

any tricks plz

Answer 1

This answer assumes you meant integer solutions.

Hint: For the first equation, $$(x-y)(x+y)=29$$ but 29 is a prime so one of both factors is equal to 1.

Answer 2

For the second equation, note that $$x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac{3}{4}y^2$$

Answer 3

Hint: For the second equation, assuming you meant real solutions. $$x^2+xy+y^2=0\Rightarrow x^3-y^3=(x-y)(x^2+xy+y^2)=0.$$ But $f(x)=x^3$ is injective. Hence, $x=y$. And thus $x=y=0$.

(Or note that $x^2+xy+y^2=(x+y/2)^2+3y^2/4$.)

Answer 4

$ x^2 - y^2 = (x+y)(x-y)$

Note this is true for all factorisations of 29, but the only one that matters is 29*1, so solve for $x+y=29, x-y=1$