# If natural numbers a, b, c, d satisfy a^2+b^2 = 41 and c^2+d^2=25 then what monic quadratic in x has zeros (a+b) and (c+d) ?

Feb 18, 2017

${x}^{2} - 14 x + 45 \text{ }$ or $\text{ } {x}^{2} - 16 x + 63$

#### Explanation:

I will assume that "natural number" includes $0$.

We may suppose that $a \le b$ and $c \le d$ since swapping $a$ and $b$ or $c$ and $d$ does not change any of the expressions ${a}^{2} + {b}^{2}$, ${c}^{2} + {d}^{2}$, $\left(a + b\right)$ or $\left(c + d\right)$.

Hence:

${a}^{2} \le \frac{41}{2} \text{ }$ so $\text{ } a = 0 , 1 , 2 , 3$ or $4 \text{ }$ and $\text{ } {a}^{2} = 0 , 1 , 4 , 9$ or $16$.

${b}^{2} = 41 - {a}^{2} = \textcolor{red}{\cancel{\textcolor{b l a c k}{41}}} , \textcolor{red}{\cancel{\textcolor{b l a c k}{40}}} , \textcolor{red}{\cancel{\textcolor{b l a c k}{37}}} , \textcolor{red}{\cancel{\textcolor{b l a c k}{32}}}$ or $25$.

${c}^{2} \le \frac{25}{2} \text{ }$ so $\text{ } c = 0 , 1 , 2$ or $3 \text{ }$ and $\text{ } {c}^{2} = 0 , 1 , 4$ or $9$.

${d}^{2} = 25 - {c}^{2} = 25 , \textcolor{red}{\cancel{\textcolor{b l a c k}{24}}} , \textcolor{red}{\cancel{\textcolor{b l a c k}{21}}}$ or $16$.

So $\left\{a , b\right\} = \left\{4 , 5\right\}$ and $\left\{c , d\right\} = \left\{0 , 5\right\}$ or $\left\{3 , 4\right\}$.

So $a + b = 9$ and $c + d = 5$ or $7$.

So the polynomial we are looking for is one of:

$\left(x - 9\right) \left(x - 5\right) = {x}^{2} - 14 x + 45$

$\left(x - 9\right) \left(x - 7\right) = {x}^{2} - 16 x + 63$