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**chiph588@** 4.) $\displaystyle b^2+(b+2)^2+(b+4)^2+1 = 3(b^2+4b+7) $

Therefore we must show $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

This is easy to see since we know $\displaystyle b $ is odd, so $\displaystyle b \equiv \pm 1 \mod{4} \Longrightarrow b^2 \equiv 1 \mod{4} $.

So we have $\displaystyle b^2+4b+7 \equiv 1+0+7 \equiv 8 \equiv 0 \mod{4} $. Hence $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

(Notice we didn't use the fact that $\displaystyle b $ was prime. This result holds for all odd integers.)