Can the sum of any 2112 consecutive positive integers be a perfect square?

no, it cannot

if n/2*(2*n+2111)=m*m            m is an integer

then  2*n^2+2111*n-2*m^2=0

and 2111+16m^2=k^2              k is an integer

 

or (k-t )*(k+t )=2111                     t=4*m

and since 2111 is a prime number

there is only one integer solution k=1056 t=1055

but t=4*m     so m cannot be an integer

qed

 

Posted by Ady TZIDON on 2007-04-05 09:24:16