Question

The sum of the square of three integers is 324. How do you find the integers?

Answer 1

The only solution with distinct positive integers is `(2, 8, 16)`

The full set of solutions is:

`{ (0, 0, +-18), (+-2, +-8, +-16), (+-8, +-8, +-14), (+-6, +-12, +-12) }`

Explanation:

We can save ourselves some effort by considering what form squares take.

If `n` is an odd integer then `n = 2k+1` for some integer `k` and:

`n^2 = (2k+1)^2 = 4(k^2+k)+1`

Notice that this is an odd integer of the form `4p+1`.

So if you add the squares of two odd integers, then you will always get an integer of the form `4k+2` for some integer `k`.

Note that `324 = 4*81` is of the form `4k`, not `4k+2`.

Hence we can deduce that the three integers must all be even.

There are a finite number of solutions in integers since `n^2 >= 0` for any integer `n`.

Consider solutions in non-negative integers. We can add variants involving negative integers at the end.

Suppose the largest integer is `n`, then:

`324/3 = 108 <= n^2 <= 324 = 18^2`

So:

`12 <= n <= 18`

That results in possible sums of squares of the other two integers:

`324 - 18^2 = 0`

`324 - 16^2 = 68`

`324 - 14^2 = 128`

`324 - 12^2 = 180`

For each of these values `k`, suppose the largest remaining integer is `m`. Then:

`k/2 <= m^2 <= k`

and we require `k-m^2` to be a perfect square.

Hence we find solutions:

`(0, 0, 18)`

`(2, 8, 16)`

`(8, 8, 14)`

`(6, 12, 12)`

So the only solution with distinct positive integers is `(2, 8, 16)`

Answer 2

`x^2+y^2+z^2=2^2 3^4 = w^2`

It is easy to show that `x,y` and `z` must be even because making `x=2m_x+1, y=2m_y+1` and `z=2m_z` we have

`4m_x^2+4m_x+4m_y^2+4m_y+4m_z^2+2=4xx 3^4` or

`2m_x^2+2m_x+2m_y^2+2m_y+2m_z^2+1 = 2 xx 3^4` which is absurd.

So we will consider from now on

`m_x^2+m_y^2+m_z^2= 3^4`

Now considering the identity

`((l^2+m^2-n^2)/n)^2+(2l)^2+(2m)^2=((l^2+m^2+n^2)/n)^2`

with `l,m,n` arbitrary positive integers and making

`{(m_x =(l^2+m^2-n^2)/n ),(m_y=2l),(m_z=2m),(m_w=(l^2+m^2+n^2)/n ):}` ------ [1]

we have

`l^2+m^2+n^2= 3^2 n` or solving for `n`

`n =1/2(9pm sqrt(9^2-4(l^2+m^2)))`

so for feasibility we need

`9^2-4(l^2+m^2)=p^2` or

`9^2-p^2=4(l^2+m^2) = q`

so for `p={1,2,3,4,5,6,7,8}` we will have

`q = {80,77,72,65,56,45,32,17}` so the feasible `q` are

`q_f = {80,72,56,32}` because `q equiv 0 mod 4`

so we have to find

`4(l_i^2+m_i^2)=q_i` or

`l_i^2+m_i^2 = 1/4 q_i = bar q_i={20,18,14,8}`

Here as we can easily verify, the only solution is for

`l_1=2,m_1=4` because

`l_1^2+m_1^2= bar q_1`

and consequently `n_1 = {4,5}`

and substituting into [1] we get

`n_1 = 4 rArr {(m_x=1),(m_y=4),(m_z =8 ):}`

`n_1 = 5 rArr {(m_x=-1),(m_y=4),(m_z =8 ):}`

giving the solution

`{(x = 2m_x = 2),(y=2m_y=8),(z=2m_z=16):}`