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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 ?

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    $\begingroup$ Assuming integer ages? $\endgroup$ Commented Dec 9, 2014 at 15:42
  • $\begingroup$ @barakmanos i suppose that was the intent, also the answer i found is in integers. $\endgroup$
    – Mike
    Commented Dec 9, 2014 at 15:44
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    $\begingroup$ If $b=0$ (all $9$ sons were born together), then you can choose any value of $a$ that you want, and $M=3a$. $\endgroup$ Commented Dec 9, 2014 at 15:57

4 Answers 4

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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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    $\begingroup$ If we take into account male life expectancy, then they probably mean the first answer. If we also take into account financial feasibility, then there are no solutions! $\endgroup$
    – amcalde
    Commented Dec 9, 2014 at 15:57
  • $\begingroup$ Financial feasibility? $\endgroup$
    – Nishant
    Commented Dec 10, 2014 at 3:48
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    $\begingroup$ @Nishant, It was a joke: Having 9 sons would be exorbitantly expensive for a 48 year old! Not to mention he would have had his eldest son at age $12$. $\endgroup$
    – amcalde
    Commented Dec 10, 2014 at 17:13
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Idea:

You can simplify things a bit if you make the middle kid be $n$ years old with a difference of $k$. Then the kids ages are $n-4k, n-3k, n-2k, n-k, n, n+k, n+2k, n+3k, n+4k$. The sum of the squares of the ages is $9n^2+60k^2$.

Dad's age is $\sqrt{9n^2+60k^2}$.

I think you can find reasonable values fairly quickly that give a realistic integer-valued solution.

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All the coefficients on the left are multiples of $3$, so $M$ must be a multiple of $3$, let $M=3N$. Now we have $a^2+8ab+\frac {68}3b^2=N^2$and $b$ must be a multiple of $3$. It is pretty unreasonable for $b$ to equal $6$, as the sons would span $48$ years, but you can try it if you want. We have still cut down the case work. Nothing more pops out at me. At this point I would just make a spreadsheet with possible values for $a$ and find which one gives a square for $N^2$

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  • $\begingroup$ You have $\frac{a^2}{3}+8ab+\frac {68}3b^2=N^2.$ $\endgroup$
    – mfl
    Commented Dec 9, 2014 at 15:56
  • $\begingroup$ @mfl: the coefficient on $a$ was $9$, so when I divide by $9$ I get $1$ $\endgroup$ Commented Dec 9, 2014 at 15:59
  • $\begingroup$ Right. I have misread $a=3.$ +1. $\endgroup$
    – mfl
    Commented Dec 9, 2014 at 15:59
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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 $b\sqrt{135}$.

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