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A man is $1$ year older than his wife is. The product of their ages is $17$ times the product of the ages of their son and their daughter. Their son is $1$ year older than their daughter is. Find the sum of all possible distinct values of the number of years in the son's age.

I have rewritten this problem exactly as it appears in a recreational math book I have, but it is driving me crazy. I feel like there isn't enough information. The answer is claimed to be $34$, but no step-by-step solution is offered any assistance would be appreciated.

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    Presumably you have to solve something like $m(m-1)=17s(s-1)$ for positive integers. To save time, note that $m$ or $m-1$ must be multiples of $17$. There is more than one solution, though one of them is slightly implausible for the mother's ages when the children were born. – 2017-01-22
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    Following @Henry you can assume the parent's ages are less than $100$. That only gives twelve choices. I find two solutions. – 2017-01-23
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    That is what I was confused about. Do I have to assume these individuals are living normal human life spans or else there would be infinite answers. I understand where the equations Henry gave came from. I just found it odd they did not mention that. – 2017-01-23

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Let's say the man's age is M, his wife's W, his son's S, and his daughter's D. Then we have $M = W + 1$, so $W = M - 1$. The product of their ages is $17$ times the product of the ages of their son and their daughter: $M(M-1) = 17SD$. $D = S - 1$. So we have $M(M-1) = 17S(S-1)$.

One significant factor is that $M$ and $S$ must be whole numbers, and $M(M-1)$ must be divisible by $17$, which means either $M$ or $M - 1$ is divisible by $17$. So $M$ must be $17, 18, 34, 35, 51, 52, 68, 69, 85, 86$ (or a few more options, depending on the man's lifespan).

The next relevant issue is divisibility by $3$. Modulo $3$, $n(n - 1)$ must be either $0$ or $2$. If $S(S-1)$ is $2$ mod $3$, then $17S(S-1)$ is $1$ mod $3$ and no value of $M$ will work; so it must be that $S(S-1)$ is divisible by $3$ (so $M(M-1)$ is as well). Now our options for $M$ are reduced to $18, 34, 51, 52, 69,$ or $85$. This is small enough to check.

$18(18 - 1) = 18 \cdot 17$, so $S(S - 1) = 18$. No integer $S$ works for this.

$34(34 - 1) = 66 \cdot 17$, so $S(S - 1) = 66$. No integer $S$ works for this.

$51(51 - 1) = 150 \cdot 17$, so $S(S - 1) = 150$. No integer $S$ works for this, either.

$52(52 - 1) = 156 \cdot 17$, so $S(S - 1) = 156$, which is satisfied if $S = 13$.

$69(69 - 1) = 276 \cdot 17$, so $S(S - 1) = 276$, which is never satisfied.

$85(85 - 1) = 420 \cdot 17$, so $S(S - 1) = 420$, which is satisfied if $S = 21$.

So our two possible ages for the son are $13$ and $21$; $13 + 21 = 34$.

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    So the mother might have been about $39$ or $64$ years old when the daughter was born. The second seems unlikely: examples from https://en.wikipedia.org/wiki/Pregnancy_over_age_50 do not seem to suggest these would have been a natural conception or that the mother also gave birth the previous year – 2017-01-23
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    @Henry In fairness, the problem never said the children were biological. These might have been adoptions. – 2017-01-23