Question

How do you solve the simultaneous equations `a^2+b^2+c^2 = 155` and `a+b+c=21` ?

Answer 1

`{a, b, c} = {5, 7, 9}`

Explanation:

Given:

`a^2+b^2+c^2 = 155`

`a+b+c = 21`

I presume you are wanting integer solutions - otherwise there are not enough constraints.

If `a, b, c` are real then `a^2`, `b^2` and `c^2` are all non-negative.

The largest of the squares must be greater than or equal to `155/3 = 51.bar(6)` and less than or equal to `155`. That gives possibilities:

`8^2 = 64`, `9^2 = 81`, `10^2 = 100`, `11^2 = 121`, `12^2 = 144`

For each of these the remainder needs to be expressed as the sum of two squares.

Enumerating the possibilities, we have:

`155-64 = 91 = 81+color(red)(cancel(color(black)(10)))`

`color(white)(155-64 = 91) = 64+color(red)(cancel(color(black)(27)))`

`color(white)(155-64 = 91) = 49+color(red)(cancel(color(black)(42)))`

`155-81 = 74 = 64+color(red)(cancel(color(black)(10)))`

`color(white)(155-81 = 74) = 49+25`

`155-100 = 55 = 49+color(red)(cancel(color(black)(6)))`

`color(white)(155-100 = 55) = 36+color(red)(cancel(color(black)(19)))`

`155-121 = 34 = 25+9`

`155-144 = 11 = 9+color(red)(cancel(color(black)(2)))`

So the only decompositions as a sum of three squares are:

`155 = 81+49+25 = 9^2+7^2+5^2`

`155 = 121+25+9 = 11^2+5^2+3^2`

We can discount the second of these since `11+5+3 = 19 != 21`, but the first has `9+7+5 = 21` as required.