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)# ?

1 Answer
Feb 18, 2017

Answer:

#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#