Case 1: m is even. Given, 8f(m+1)−f(m)=2 ⇒8(m+1+3)−m(m+1)=2 ⇒8m+32−m2−m=2 ⇒m2−7m+30=0 ⇒(m−10)(m+3)=0 ⇒m=10‌ or ‌−3 As m is positive integer, the only possible value of m=10. Case 2: If m is odd, then we would not be getting positive solution.