Answer:
yes
Step-by-step explanation:
Let's call the first integer n and let's compute.[tex]n^2 + (n+1)^2 + (n+2)^2 +(n+3)^2 = \\n^2 + \\n^2+2n+1 + \\\ n^2+4n+4 + \\\ n^2+6n+9+\\\ n^2+8n+16 =\\4n^2+20n +30 = 2 (2n^2+10n+15)[/tex]
Which is indeed of the form 2(something), so it is even.
