Question

If a, b, c and d are natural numbers such that a²+b² = 41 and c² + d² = 25, then find the polynomial whose zeroes are (a + b) and (c+ d)

Answer 1

since a,b,c,d are natural numbers therefore
${a}^{2} + {b}^{2} = 41$
is only possible when
$a = 5 \\ b = 4$
similarly for c and d
$c = 4 \\ d = 3$
so the equation with whose roots are
$a + b = 9 \: \\ c + d = 7$
is
${x}^{2} - 16x + 63$