+0  
 
0
91
2
avatar+169 

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

matth99  Jun 29, 2018
 #1
avatar+142 
+1

900\( \)

DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
 #2
avatar+88898 
+1

Let's see how Daniel might have gotten this answer.....

 

Let  a  > b   where  a and b are positive integers

 

Then  ....  .let  b^2  and a^2  be consecutive squares

 

So

 

a^2  -  b^2  =  59

( a + b) ( a - b)  = 59

 

Since   a, b are integers  and  59 is prime...its  only divisors  are 1  and 59

 

This implies that

 

a + b   =  59

a - b  = 1           add these

 

2a  = 60

a  = 30

 

And

a - b  =1

a - 1   = b

30 - 1  =  29

 

So

 

30^2  - 29^2  =

900  - 841  =

59

 

So  900  is the larger perfect square

 

 

cool cool cool

CPhill  Jun 29, 2018

34 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.