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Recall that a perfect square is the square of some integer. How many perfect squares less than 10,000 can be represented as the difference of two consecutive perfect cubes?

 Feb 12, 2021
 #1
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Every perfect square which is an odd number can be represented as the difference of two consecutive perfect squares. The odd numbers less than 100 (which is the square root of 10000) should be the answer. So, the answer is 50

Explanation:

 

Consider (x+1)^2 and x^2 be two consecutive perfect squares.

Difference of those two:

(x+1)^2 - x^2

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

= 2x + 1 = is an odd number and should also be a perfect square.

For example: consider an odd perfect square : 25

Which can be written as 2(12)+1here, x=12

25 can be written as (x+1)^2 - x^2

= (12+1)^2 - 12^2

= 13^2 - 12^2

= 169 - 144

= 25

 

I hope you understood.
 

 Feb 12, 2021
 #2
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The question is about a "perfect square that can be represented as the difference of two consecutive PERFECT CUBES" !?

Guest Feb 13, 2021

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