(2^2+4^2+…+(2n)^2) – (1^2+3^2+…+(2*n-1)^2) = k^2

 
 
The sum of the squares of the first n even integers:
2^2+4^2+6^2+...+(2 \, n)^2   ……………….   (1)

The sum of the squares of the first n odd integers
1^2+3^2+5^2+...+(2 \, n - 1)^2   ……………   (2)

Let’s find n so that   (1) – (2)   is a square
(2^2+4^2+6^2+...+(2 \, n)^2) - (1^2+3^2+5^2+...+(2 \, n - 1)^2) = k^2

 
Here are the first few solutions:
 

(2^2+4^2+6^2+8^2) - (1^2+3^2+5^2+7^2) = 6^2

(2^2+4^2+6^2+...+288^2) - (1^2+3^2+5^2+...+287^2) = 204^2

(2^2+4^2+6^2+...+9800^2) - (1^2+3^2+5^2+...+9799^2) = 6930^2

(2^2+4^2+6^2+...+332928^2) - (1^2+3^2+5^2+...+332927^2) = 235416^2

(2^2+4^2+6^2+...+11309768^2) - (1^2+3^2+5^2+...+11309767^2) = 7997214^2

(2^2+4^2+6^2+...+384199200^2) - (1^2+3^2+5^2+...+384199199^2) = 271669860^2

(2^2+4^2+6^2+...+13051463048^2)-(1^2+3^2+5^2+...+13051463047^2)=9228778026^2

(2^2+4^2+6^2+...+443365544448^2)-(1^2+3^2+5^2+...+443365544447^2)=313506783024^2

(2^2+4^2+6^2+...+15061377048200^2)-(1^2+3^2+5^2+...+15061377048199^2)=10650001844790^2

(2^2+4^2+6^2+...+511643454094368^2)-(1^2+3^2+5^2+...+511643454094367^2)=361786555939836^2
 
 
write a recursion formula

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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