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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?

 Aug 2, 2021
 #1
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Note that...

 

$1-0 = 1$

$4-1 = 3$

$9-4 = 5$

$16-9 = 7$

$25-16 = 9$

$...$ 

 

Thus, every positive odd number can be written as the difference between two perfect squares. We consider squares in the range of $1 - 99$, since $100^2 = 10,000$. An even number squared will never be odd, and vice versa an odd number squares will always be even. 

 

Thus, taking $\frac{99+1}{2} = \boxed{50}$, which is our answer. 

 Aug 2, 2021
 #2
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That's interesting, but it isn't what the problem was asking for. 

 

I'm not sure what the problem wants but I think it's  (n + 1)3 – (n)3  =  x2  

Guest Aug 2, 2021

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