Can you find it?
Find the second least positive integer such that is the sum of squares of consecutive integers.
Details and Assumptions
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The least value is where the consecutive integers start from , i.e.
The answer is 2161.
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1 solution
If m 3 = ( x + 1 ) 2 + ⋯ + ( x + m ) 2 , then 4 m 2 − 3 m ( 2 x + 1 ) − ( 6 x 2 + 6 x + 1 ) = 0 and the discriminant of this polynomial is 3 3 ( 2 x + 1 ) 2 − 8 . It has to be a square, so we're looking for integer solutions to y 2 − 3 3 z 2 = − 8 where z = 2 x + 1 .
In the interest of conciseness, I'll spare you the Pell equation derivation. The first three solutions with y , z positive are ( y , z ) = ( 5 , 1 ) , ( 2 4 7 , 4 3 ) , ( 1 1 3 5 7 , 1 9 7 7 ) , which lead to ( m , x ) = ( 1 , 0 ) , ( 4 7 , 2 1 ) , ( 2 1 6 1 , 9 8 8 ) respectively. So the answer is 2 1 6 1 .