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The positive difference between two consecutive perfect squares is 59. What is the greater of the two perfect squares?

 Feb 8, 2019

Hello, APatel!


A perfect square is the product of a whole number multiplied by itself. Below is a list of the first few perfect squares, in order.


x x^2
1 1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81


Let equal a positive integer. Let x^2 equal any perfect square. 

Let x+1 equal the subsequent positive integer. Let (x+1)^2 equal the next perfect square.


Notice how this is very similar to how I generated consecutive perfect squares in the table; I incremented the "x" column by 1 and squared the result. Also, notice that the question asks for the positive difference of the perfect squares. \((x+1)^2>x^2\) because (x+1)^2 represents the next perfect square. Therefore, it is possible to generate the following equation for this problem:

\((x+1)^2-x^2=59\) Expand the binomial.
\(x^2+2x+1-x^2=59\) Look at that! The quadratic terms cancel out, and we are left with quite a basic equation.
\(2x+1=59\) Now, subtract one on both sides.
\(2x=58\) Divide by 2 on both sides.


We are not done yet! Remember, our goal is to find the greater of the perfect squares. Of course, in this problem, I let (x+1)^2 represent the larger one.


\(x=29\\ (x+1)^2\) Substitute 29 into this expression and simplify accordingly. 
\(900\) This is the larger of the 2 perfect squares. 


 Feb 8, 2019

\(\text{let }a>b\\ a^2-b^2 = 59\\ (a+b)(a-b)=59\\ \text{The only factors of 59 are 1 and 59}\\ (a+b)=59\\ (a-b)=1\\ 2a=60\\ a=30\\ b=29\\(30)^2=900 \text{ is the larger of the two squares}\)

 Feb 8, 2019
edited by Rom  Feb 8, 2019

A lot of times, you can eliminate a step by assigning 'x' the value you are looking for....


x = larger number

x-1 = the previous number


The difference in these squares = 59


x^2 - (x-1)^2 = 59     expand

x^2 - x^2 + 2x -1 = 59

2x-1 = 59

2x = 60

x = larger number = 30         x^2 = 900

 Feb 8, 2019

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