TITLE: Solving 9 sons puzzle
QUESTION [16 upvotes]: The following math puzzle : 
There is a person with 9 sons, 
they all are separated by equal amount of years (equal intervals). 
if the age of the person squared is the same as 
the sum of the squares of his own kids.
how old are his kids.

I could only solve this brute force. was looking to see if there is an analytic way of approaching this.
where I got to:

$a$ = first boys age
$b$ = interval
$M$ = The fathers age
$$\sum _{k=0}^8 (a+b k)^2=9 a^2+72 a b+204 b^2 = M^2$$

from here I can only go brute force... which yields $a=2$, $b=3 \implies M=48$
does anyone know how to solve this with out going over all values ?

REPLY [3 votes]: The discriminat of the equation
$$9a^2+72ab+204b^2-M^2=0$$
is $$4(M^2-540b^2)$$
so $M^2-540b^2$ must be a perfect square, say, $n^2$. Therefore,
$$540b^2=M^2-n^2=(M+n)(M-n)$$
The arithmetic mean of $M+n$ and $M-n$ is $M$, and it is greater that the geometric mean, $\sqrt{540b^2}>23b$. Since $M$ is the age of a man, $b\le 5$.
Since $M+n$ and $M-n$ have the same parity, both must be even, so we can write
$$135b^2=\frac{M+n}2\cdot\frac{M-n}2$$
and $\dfrac{M-n}2$ is a divisor of $135b^2$ lesser than $b\sqrt{135}$.