Question

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)` ?

Answer 1

`x^2-14x+45" "` or `" "x^2-16x+63`

Explanation:

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

We may suppose that `a <= b` and `c <= 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`, `(a+b)` or `(c+d)`.

Hence:

`a^2 <= 41/2" "` so `" "a = 0, 1, 2, 3` or `4" "` and `" "a^2 = 0, 1, 4, 9` or `16`.

`b^2 = 41 - a^2 = color(red)(cancel(color(black)(41))), color(red)(cancel(color(black)(40))), color(red)(cancel(color(black)(37))), color(red)(cancel(color(black)(32)))` or `25`.

`c^2 <= 25/2" "` so `" "c = 0, 1, 2` or `3" "` and `" "c^2 = 0, 1, 4` or `9`.

`d^2 = 25-c^2 = 25, color(red)(cancel(color(black)(24))), color(red)(cancel(color(black)(21)))` or `16`.

So `{a, b} = {4, 5}` and `{c, d} = {0, 5}` or `{3, 4}`.

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

So the polynomial we are looking for is one of:

`(x-9)(x-5) = x^2-14x+45`

`(x-9)(x-7) = x^2-16x+63`