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Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.

Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?

Here's what I did:

Let the population last year be n, so n = x^2 and x = √n Last month: n + 100 = x^2 + 1 Next Month: n + 200 = x^2 ...

and i Got stuck there. I don't know where I am going ... Your help is appreciated

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I like where Will Jagy starts.

$$x^2+99 = y^2\\ x^2 + 200 = z^2$$

to continue I would subtract one from the other

$$z^2 - y^2 = 101\\(z+y)(z-y) = 101$$

$101$ is prime

$$z+y = 101\\z-y = 1\\z = 51\\y = 50\\x = \sqrt{51^2 - 200} = 49$$

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  • $\begingroup$ good........... $\endgroup$ – Will Jagy Jan 30 at 0:43
  • $\begingroup$ on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :( $\endgroup$ – harpey1111 Jan 30 at 22:34
  • $\begingroup$ and maybe that 50^2 is supposed tobe 51^2 right? $\endgroup$ – harpey1111 Jan 30 at 22:45
  • $\begingroup$ @harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out. $\endgroup$ – Doug M Jan 30 at 23:08
  • $\begingroup$ thank you so much for your help. I got the problem right. have a wonderful night :) $\endgroup$ – harpey1111 Jan 30 at 23:34
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Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!

Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?

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  • $\begingroup$ I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry. $\endgroup$ – harpey1111 Jan 29 at 23:57
  • $\begingroup$ thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful. $\endgroup$ – harpey1111 Jan 30 at 23:38
  • $\begingroup$ You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best. $\endgroup$ – rogerl Jan 31 at 14:06
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You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.

It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.

Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2\cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2\cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.

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  • $\begingroup$ thank you for spending some of your time explaining this. It helped me. I really appreciate it. $\endgroup$ – harpey1111 Jan 30 at 23:39
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$$ x^2 + 99 = y^2 $$ $$ x^2 + 200 = z^2 $$ $$ (y+x)(y-x) = 99 = 99 \cdot 1 = 33 \cdot 3 = 11 \cdot 9$$ The choices for $x$ are $$ \frac{99-1}{2} , \; \; \; \frac{33-3}{2} , \; \; \; \frac{11-9}{2} , \; \; \; $$ Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.

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  • $\begingroup$ how did you get "divided by 2"? $\endgroup$ – harpey1111 Jan 30 at 0:32
  • $\begingroup$ @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$ $\endgroup$ – Will Jagy Jan 30 at 0:37
  • $\begingroup$ Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :) $\endgroup$ – harpey1111 Jan 30 at 23:40

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