Question 136433
In order to get rid of any {{{i}}} in the result, the factors
would have to be of the form {{{a + bi}}} and {{{a - bi}}}, then
{{{(a + bi)(a - bi) = a^2 - b^2*(-1)}}}
{{{a^2 - b^2*(-1) = a^2 + b^2}}}, and it's given that
{{{a^2 + b^2 = 17}}}
I know that {{{16}}} and {{{1}}} are both squares and add up to {{{17}}}, so
{{{4^2 + 1^2 = 17}}}
{{{a = 4}}}
{{{b = 1}}}
So, the numbers are
{{{4 + i}}} and
{{{4 - i}}}