Tricky Problem

Ok so you are given; a, b, c are integers

abc = 729
and a + b + c = 91

Find a^2 + b^2 + c^2

How could I do this algebraically?

Hint: 1+9+81=91

Reply

Say $c = 1$ . Therefore, we have :

$ab = 729$
$a + b = 90$

Substitute :

$a(90 - a) = 729$
$90a - a^2 = 729$
$a^2 - 90a + 729 = 0$

Solve for $a$ .

Facts for the lazy :

Spoiler:

You sure ?

Spoiler:

$\triangle = 90^2 - 2916 = 5184$

$a = \frac{\ {90 - \sqrt{\triangle}}}{2} = 9$

$b = \frac{\ {90 + \sqrt{\triangle}}}{2} = 81$

And for the curious :

Spoiler:

Since you would get a polynomial with the same coefficients if you had substituted $b$ instead of $a$ , the two solutions of the quadratic equation are $a$ and $b$ .
Oh, and by the way, you just factored 729 into smaller factors :p

Finding the sum of the squares shouldn't be too hard now ;) (remember $c = 1$ )