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A very simple problem tends to become very hard. Perhaps I am overthinking it.

If the square of Winslow's age is added to Abby's age, the sum is 209. If the square of Abby's age is added to Winslow's age, the sum if 183. What is the sum of their ages?

I start off as usual by setting equations:

$$w^2+a=209$$ $$w+a^2=183$$

where $w$ and $a$ are Winslow's and Abby's ages, respectively.

Now, I try to solve:

$$w=183-a^2$$$$\implies (183-a^2)^2+a=209$$$$\implies a^4-366a^2+a+33280=0$$

I have no way to continue solving, as I can't set $a^2$ to some variable due to there being an $a$ in the equation. I try to solve it again by substituting for $w$ but still get the very complicated equation.

Desperate, I try adding the equations together:

$$\implies w^2+w+a^2+a=392$$$$\implies w(w+1)+a(a+1)=392$$

Here, I become stuck again, as I cannot factor this thing further into the form $(a+b)(c+d)$.

How do I solve this problem? Have I missed a point?

Thanks!

Max0815

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    $\begingroup$ This may not help but I did it by inspection and got $w=14$ and $a=13$. $\endgroup$ – John Douma Mar 21 at 21:41
  • $\begingroup$ Inspection may be the best way to go here. $\endgroup$ – Minus One-Twelfth Mar 21 at 21:42
  • $\begingroup$ Hint? The question asks for the sum of the ages, not the ages themselves. I think if you add the equations and subtract them you will be on your way. $\endgroup$ – Ethan Bolker Mar 21 at 21:42
  • $\begingroup$ @EthanBolker I'm not sure how adding the equation and then subtracting it again will do any help as I currently understand your hint, because doing that will just reverse the adding. $\endgroup$ – Max0815 Mar 21 at 21:47
  • $\begingroup$ @Matt I see. Mind to write an answer? $\endgroup$ – Max0815 Mar 21 at 21:54
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Instead of adding the equations, you could subtract them:

$209-183=w^2+a-a^2-w=(w+a)(w-a)+(a-w)=(w-a)(w+a-1).$

Since $209-183=26=13 \cdot 2=1\cdot 26$, and $w+a-1 > w-a$, we have that either $w+a-1=26$ or that $w+a-1=13$. But since $w$ and $a$ must have different parity (if they had the same parity, $w^2+a$ and $a^2+w$ would be even, which is not the case), their sum must be odd, meaning that we can rule out the $w+a=14$ case. Therefore, $w+a=27$.

(I have edited this comment, I dont know if that is something that I am supposed to mention, but now I have).

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Ages must be positive whole numbers, so the sums give us bounds:

$a^2$ must be less than $183$, so $a$ $<$ $14$.

$w^2$ must be less than 209, so $w$ $<$ $15$.

Testing the highest possible pair gives us a solution:

$$14^2+13=209$$

$$13^2+14=183$$

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