Case 1) If m is odd, then m+1 will be even

So f(m) = m+3 and f(m+1) = m(m+1)

Given, 8f(m + 1) – f(m) = 2

Or 8* (m+2)(m+1)- (m+3) = 2

8m^2 + 24m + 16 – m – 3 =2

8m^2 + 23m – 11 = 0

No integer solution so m is not odd means m is even.

Case 2) if m is even , m+1 = odd

So f(m+1) = m+3 and f(m) = m(m+1)

Given, 8f(m + 1) – f(m) = 2

Or 8(m+1 + 3) – m(m+1) =2

8m + 32 – m^2 –m =2

m^2 -7m -30 =0

(m+3)*(m-10)=0

M = 10 or -3

As m is even so m =10