Equations: Find $c,b,f$ if $c,b,f>0$

Question

I am given $c^2+f^2+cf=49$, $c^2+b^2-cb=49$ and $f^2+b^2-fb=49$. Find $c,b,f$ if $c,b,f>0$

I couldn't do this by hand, please help

All I can find out is that $c+f=b$.

Answer 1

We have

$$c^2+f^2+cf=49,$$ $$c^2+b^2-cb=49,$$ $$f^2+b^2-fb=49.$$

Subtracting the second equation from the first, we get

$$f^2 - b^2 + cf + cb = 0$$

Factoring,

$$(f+b)(f-b) + c(f+b) = 0$$

Factoring further,

$$(f+b)(c + f - b) = 0$$

So we have either $b = -f$ or $b = c + f$. We'll need to handle these separately.

Case 1: Suppose $b = -f$. Then the system becomes

$$c^2+f^2+cf=49,$$ $$f^2=49,$$ $$b = -f.$$

From the second equation, either $f = -7$ or $f = 7$. In the former case, the first equation becomes $c^2 - 7c = 0$, and in the latter it becomes $c^2 + 7c = 0$. So we have four solutions all together:

$$(7, 0, -7), (7, -7, -7), (-7, 0, 7), \text{ and } (-7, -7, 7).$$

Case 2: Suppose $b = c + f$. Then the system becomes

$$c^2 + cf + f^2 = 49,$$ $$b = c + f.$$

This is a one-dimensional infinite family of solutions.

So, all together, the solutions to this equation form a curve together with four isolated points in three-dimensional space. If you want integer solutions only, it's easy to see that there are 18 integer points on the ellipse in case 2, for a total of 22 integer solutions to the equation.