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

##### 2 Answers

The only solution with distinct positive integers is

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^2 = (2k+1)^2 = 4(k^2+k)+1#

Notice that this is an odd integer of the form

So if you add the squares of two odd integers, then you will always get an integer of the form

Note that

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

There are a finite number of solutions in integers since

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

Suppose the largest integer is

#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/2 <= m^2 <= k#

and we require

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

It is easy to show that

So we will consider from now on

Now considering the identity

with

we have

so for feasibility we need

so for

so we have to find

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

and consequently

and substituting into [1] we get

giving the solution