There is actually a proof for the pattern. I'll post something more on this when I have time later.

EDIT:
OK, I have some time now so here is a bit more detailed explanation.

As noted, the amount of money is a perfect square, with its tens digit being odd.

Method 1:

Some properties of perfect squares:

All perfect squares end in 1, 4, 5, 6, 9 or 00.
If it ends in 6, ten’s digit is always odd (1, 3, 5, 7, and 9), otherwise, if it ends in 1, 4, 5, or 9 the ten’s digit is always even (2, 4, 6, 8, 0).

That is enough to solve the problem. For those interested, there are more properties. You can look for proof and more properties on the web. It would be too cluttered to post it here. From Wikipedia:

If the last digit of a number is 0, its square ends in an even number of 0s (so at least 00) and the digits preceding the ending 0s must also form a square.
If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four.
If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even.
If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four.
If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd.
If the last digit of a number is 5, its square ends in 25 and the preceding digits must form a pronic number.

Method 2:

With a "proof" of mine. It might not be too neat or original, there are probably better ones out there. But here goes.

Hypothesis: perfect square mod 20 gives one of the following: 0, 1, 4, 5, 9 or 16

"Proof":

Any natural number can be expressed in the following form:
10a + b
where a is a non-negative integer and b is a non-negative integer smaller than 10

e.g.
28 = 2 x 10 + 8
927 = 92 x 10 + 7
5 = 0 x 10 + 5
... etc

If we square it
(10a + b) ^ 2 = 100a^2 + 20ab + b^2

100a^2 is divisible by 20
20ab is divisible by 20

So we're left with only b^2 where b is in range 0-9
The possible values of b^2 are: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81
All these values mod 20 give one of the following: 0, 1, 4, 5, 9 or 16.

Hence proven.

Of these values, we have to disregard the ones smaller than 10, since they were left with an odd 10$ bill and loose change. Therefore this had to be 16. The elder brother took the extra 10$ bill, while the younger one took 6$ and the bag, which had to be worth 4$.

Doesn't have to be strictly maths. See if you can come up with anything. If you still have nothing, pass your turn to whoever wants to take it.