Htykuut

I've got this system of equations.
(a,b,c) ∈ ℤ

$a*b+1 = c$

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

$2a + b = c$

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?

Thx

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 $a$.

Eliminating $c$ from the first two equations,

$$(a-b)^2=1$$ or $$a=bpm1.$$

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

$$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$

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

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, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
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.