Atown, Where Everyone Is Square

Let the population in Dec 1900 be $x^2$ . Then population in Jan 1991 can be written as $x^{2} + 48 = n^2 + 1$ . Here $x$ and $n$ are non-negative integers. We get $(n+x)(n-x) = 47$ . The only positive divisors of $47$ are $1$ and $47$ . So, one of $n+x$ , $n-x$ is $47$ and other one is $1$ . Since, $n+x \geq n - x$ , we get $n+x = 47$ and $n - x = 1$ . Solving these, we get $x = 23$ and $n = 24$ . Now we check the third condition. Population in Feb 1991 comes out to be $x^2 + 96 = 625 = 25^2$ . So, the initial population is $529$ .

Let $a^2$ be population in Dec 1900, $b^2+1$ in Jan 1901, $c^2$ in Feb 1901. So $c^2 - b^2 = 49 (c+b)(c-b) = 49 = 7^2$ . Since $b \neq 0$ , $c-b = 1$ , $c+b =49$ . Since $c>b$ , $c=25$ , $b= 24$ . $a^2+48 = 24^2+1$ . Thus $a^2= 529$ , which is the required answer.

Let x=a^2 be the poputation in December 1900, b^2+1 be the population in Januaray 1901 and c^2 be the population in February 1901. Then we have:

(1) x=a^2; (2) x+48=b^2+1 and (3) x+96=c^2.

By combining (1) and (2), we have:

b^2-a^2=(b+a)(b-a)=47.

Since 47 can only be factorized into 47*1, given a and b are non-negative, we conclude that:

a+b=47 and b-a=1, which leads to a=23, b=24, c=25;

So that the population of Atown in December 1900 is 23*23=529.

let x^2 be population of Atown. Then we need to solve, y^2-x^2=47 and z^2-y^2=49 . Since x,y,z all are positive integers so y^2-x^2=47 implies x=23 and y=24, as 47 is prime.

we get x^2=529, population of Atown.

Let the population of the city be x.

Then, x + 48= y^2 + 1 for some positive integer y. y + 48 + 48= z^2 for some positive integer z.

So, x + 47= y^2 .....(1) y + 96= z^2 .....(2)

Subtract equation (1) from equation (2)...

z^2 - y^2= 96 - 47 Or, (z+y)(z-y)= 49

Note that z+y can never be negative, so neither can be z-y.

Now, 49 can be factored into positive integers in two ways:- 49= 7 7= 49 1

Consider the first case.

We have, z+y= 7, z-y= 7. Then, y=0 and x= -96, which is impossible since the population of a town can never be negative.

So, the second case must be true.

Since z+y>z-y, z+y= 49, z-y= 1. Solving, we get z= 25, y= 24.

Then, z= z^2 - 47 = 25^2 - 47= 529

So 529 is the answer.

Let N be the population in December. Then we are looking for N such that N is a perfect square; N+47 is a perfect square, and N+96 is a perfect square.

By inspecting a list of perfect squares, we find that N = (23)^2 = 529 satisfies all of the conditions.

suppose the population on 1900 = x^2 , so x^2 + 48 = 1 + y^2 1 + y^2 + 48 = n^2 so, x^2 + 96 = n^2 we got (n + x)(n-x)=96, then we got n = 25, y=24, and x = 23 the population on 1900 is x^2 = 23^2 = 529

let us assume that few things: Population in 1990 = x^2 Population in 1991 Jan = x^2 + 48 = y^2+1 (one more than a perfect square) ----------(1) Population in 1991 Feb = x^2 + 96 = z^2 (perfect square)--------------------------------------(2) Subtract (2) - (1) Z^2 - Y^2 -1 = 48 Z^2 - y^2 = 49 (z- y) (z + y) = 49-----------------------------------------------------(3) analyzing RHS of equation 3 49 = 1*49 = 7 *7 (factors of 49) (z- y) (z + y) = (7) * (7) ------ this is not possible for obvious reasons... sum and difference of two numbers cannot be same until one of them is 0....and here none can be 0 as they denote population which is definitely non zero.... :) so (z- y) (z + y) = (1) * (49) hence (z- y) = 1 (z + y) = 49 solve z= 25 y = 24 and when we substitute their values to any of equation 1 or 2 we get x^2 = 529...answer

the correct ans is 529... month wise population is Dec'1900 = 529, Jan'1901 = 577 and Feb'1901 = 625

Here, The right answer according to the question is 625.

But according to answer given by you there shold be a change in the question that is you should ask the population of atown in February 1901.

So please check and correct it......