[Math] Solving 9 sons puzzle

binomial theorempuzzlerecreational-mathematics

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 ?

Best Answer

In your equation, if you solve for $a$ you get:

$$a = \frac{\sqrt{M^2-60b^2}}{3} - 4b$$ If we're looking for integer solutions then this means that $M^2-60b^2$ must be a square. Also it must be divisible by $3$, so at least you only need search over (pseudo-)triples of the form: $$9x^2 + 60b^2 = M^2$$ Moreover you want $a$ to be positive so you require $x \ge 4b$. There is only one primitive triple with this property which yields $\{a,b,M\} = \{2,3,48\}$. Any multiple of this triple would also work ($\{4,6,96\}, \{6,9,144\}, \ldots$).

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