Solve over the integers

Question

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

Maybe $x = \frac{n(n-1)}2$ and $y = \frac{n(n+1)}2$

2) Find all integers solution in system of two equations: $$8x+5y+z=100 \quad \text{ and }\quad x+y+z=20$$

3)$\frac1x+\frac1y+\frac1z = 1$

I tried to arrange the variables but it didn’t lead to an explicit answer

Answer 1

$(1.)$

$$x+y= (x-y)^2 \implies x^2 - x(2y+1) + (y^2-y) = 0$$ Using quadratic formula , we get :

$$x = \frac{(2y+1) \pm \sqrt{4y^2 + 1+4y - 4y^2 +4y}}2 \\ x = \frac{(2y+1) \pm \sqrt{8y+1 }}2 $$

Since $x\in \mathbb{Z}$ , we have that $(8y+1)$ is a perfect square. Taking $y = (2n^2-n)$_{$\color{magenta}{[1.]}$} for some $n\in\mathbb{N}$ , we have $8y+1 = (4n-1)^2$

Hence $$x = \frac{(2(2n^2-n)+1) \pm (4n-1)}2$$ $$\color{#d07}{x = 2n^2+n} \quad\text{ or } \quad \color{#3d0}{x = 2n^2 - 3n +1}$$

Hence possible solutions are $(x,y) = \color{#25f}{(2n^2+n,2n^2-n)}$ or $\color{#25f}{(x,y) = (2n^2-3n+1 , 2n^2-n)}$

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$\color{magenta}{[1.]}$ On clarification on how I got $y = 2n^2 - n$ , I realized that to make $8y+1$ a perfect square , I wanted it to be of the form $(an-1)^2$ or $a^2n^2 - 2an + 1$. So $y$ should be of form $bn^2 - cn$ , which gives us $8y+1 = 8bn^2 - 8cn + 1$ . To make this a perfect square , I took $b = 2$ , which makes $8y+1 = 16n^2 - 8cn + 1$. Now if we compare it with $(4n-1)^2$ , we get $c=1$ . And from there , I deduced that $\boxed{y = 2n^2-n}$

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$(2.)$

$$8x+5y+z=100 \\ x+y+z=20$$ Subtracting both the equations , we get :

$$7x + 4y = 80 \implies 4y - 80\equiv 0 \mod 7$$ $$4y \equiv 3 \mod 7 \implies \color{#d05}{y = 6 + 7n}$$ Hence $$7x+24 + 28n = 80 \implies \color{#3d0}{x= 8-4n}$$

And $$8-4n + 6 + 7n + z = 20 \implies \color{#30a}{z = 6-3n}$$

All the solutions are given by $\boxed{\color{#20f}{(x,y,z) = (8-4n\,,6+7n\,,6-3n)}}$

Answer 2

Probably it is necessary to draw a formula for the solution in the general form:

In the equation: $$(x-y)^2=x+y$$

Solutions can be written:

$$x=\frac{a(a+1)}{2}+\frac{b(b-1)}{2}-ab$$

$$y=\frac{a(a-1)}{2}+\frac{b(b+1)}{2}-ab$$

$a,b$ - what some integers.