Some of square numbers

Number Theory Level pending

Which of the following statements is/are true?

\quad A : It is possible that if k k is a positive integer, then k 2 + ( k + 1 ) 2 + ( k + 2 ) 2 + ( k + 3 ) 2 + ⋯ + ( k + 8 ) 2 = prime number k^2+(k+1)^2+(k+2)^2+(k+3)^2+\dots+(k+8)^2=\text{prime number}

\quad B : It is possible that if k k is a positive integer, then k 2 + ( k + 1 ) 2 + ( k + 2 ) 2 + ( k + 3 ) 2 + ⋯ + ( k + 8 ) 2 = perfect square k^2+(k+1)^2+(k+2)^2+(k+3)^2+\dots+(k+8)^2=\text{perfect square}

Both of A and B Neither A nor B Only A Only B

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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1 solution

Áron Bán-Szabó
Jul 30, 2017

Let k + 4 = n k+4=n . Then

( n − 4 ) 2 + ( n − 3 ) 2 + ( n − 2 ) 2 + ( n − 1 ) 2 + n 2 + ( n + 1 ) 2 + ( n + 2 ) 2 + ( n + 3 ) 2 + ( n + 4 ) 2 = 9 n 2 + 2 ∗ ( 1 2 + 2 2 + 3 2 + 4 2 ) = 9 n 2 + 60 = 3 ( 3 n 2 + 20 ) (n-4)^2+(n-3)^2+(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2=9n^2+2*(1^2+2^2+3^2+4^2)=9n^2+60=3(3n^2+20)

Since 3 n 2 + 20 ≠ 0 3n^2+20\neq 0 , it can't be a prime number, because 3 3 is a divisor of it. Note that if a perfect square is divisible by q q , where q q is a prime number, then it is divisible by q 2 q^2 . But this condition is not true for 3 ( 3 n 2 + 20 ) 3(3n^2+20) , because 3 ∤ 3 n 2 + 20 3\not\mid 3n^2+20 (because 3 ∤ 20 3\not\mid 20 ).

Therefore neither them is possible.

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