Answer to Question #167780 in Math for ZXCVC

Question

4. The population of a certain village at one time was a perfect square. Later, with an increase of 100, the population was one more than a perfect square. Now, with an additional increase of 100, the population is again a perfect square. What was the original population? (Give all possible answers.)

Answer 1

Let $n^2=$ original population count. Then

$$n^2+100=a^2+1$$

$$n^2+200=b^2$$

$$(a-n)(a+n)=99$$

$$(b-n)(b+n)=200$$

$$99=1\times99=3\times33=9\times 11$$

$$200=1\times200=2\times100=4\times50=5\times20$$

$$=8\times25=10\times20=8\times25$$

$$\begin{matrix} a-n=1 \\ a+n=99 \end{matrix}=>\begin{matrix} n=49 \\ a=50 \end{matrix}$$

$49^2=2401,$

$49^2+100=2501=50^2+1,$

$49^2+200=2601=51^2$

$$\begin{matrix} a-n=9 \\ a+n=11 \end{matrix}=>\begin{matrix} n=15 \\ a=18 \end{matrix}$$

$15^2=225,$

$15^2+100=325=18^2+1,$

$15^2+200=425, 20^2=400, 21^2=441$

Does not satisfy.

$$\begin{matrix} a-n=1 \\ a+n=99 \end{matrix}=>\begin{matrix} n=1 \\ a=10 \end{matrix}$$

$1^2=1,$

$1^2+100=101=10^2+1,$

$1^2+200=201, 14^2=196, 15^2=225$

Does not satisfy.

The original population of a certain village was 2401 people.