# How To solve This Perfect Square Word Problem

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

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?

• 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. Jan 29, 2019 at 23:57
• thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful. Jan 30, 2019 at 23:38
• You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best. Jan 31, 2019 at 14:06

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

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

• good........... Jan 30, 2019 at 0:43
• on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :( Jan 30, 2019 at 22:34
• and maybe that 50^2 is supposed tobe 51^2 right? Jan 30, 2019 at 22:45
• @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. Jan 30, 2019 at 23:08
• thank you so much for your help. I got the problem right. have a wonderful night :) Jan 30, 2019 at 23:34

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 $$51^2$$.

• thank you for spending some of your time explaining this. It helped me. I really appreciate it. Jan 30, 2019 at 23:39

$$x^2 + 99 = y^2$$ $$x^2 + 200 = z^2$$ $$(y+x)(y-x) = 99 = 99 \cdot 1 = 33 \cdot 3 = 11 \cdot 9$$ 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.

• how did you get "divided by 2"? Jan 30, 2019 at 0:32
• @harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$ Jan 30, 2019 at 0:37
• Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :) Jan 30, 2019 at 23:40