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

 Jun 29, 2018
 #1
avatar+143 
0

900\( \)

.
 Jun 29, 2018
edited by DanielCai  Jun 29, 2018
 #2
avatar+98196 
+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

 Jun 29, 2018

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