Equation solving?

Question

I've got this system of equations: $(a,b,c)\in\mathbb{Z}$

$ab+1 = c$

$a^2 + b^2 +1 = 2c$

$2a + b = c$

I tried to substitute a little bit:

$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$

Ultimately:

$a(a-2-b) + b(b-1) = 0.$

Now I'm not sure how to substitute b into a in order to go on.

Maybe I didn't even start well. Can someone help?

Answer 1

Eliminating $c$ from the first two equations,

$$(a-b)^2=1$$ or $$a=b\pm1.$$

Then eliminating $a$ and $c$ from the first and third, we get two quadratic,

$$(b+1)b+1=2(b+1)+b\\\to b=1\pm\sqrt2,a=2\pm\sqrt2,c=5\pm3\sqrt2,$$

$$(b-1)b+1=2(b-1)+b\\\to b=1,3, a=0,2, c=1,7.$$

Answer 2

Here is a possible way to solve your problem.

From the first and second equation, we have, $$a^2+b^2+1=2ab+2\\or,(a-b)^2=1\\or,a-b=\pm1$$ Now, using 3rd equation we get, $$a=\frac{1}{3}(c\pm 1)$$ So, $$b=a\mp1=\frac{1}{3}(c\mp 2)$$ I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $\mp=-(\pm)$.

Hope it works.

Answer 3

You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.