Question 843158
What you meant to say is (x+1)(x+2)(x+3)(x+4) + 1 = n^2. A brute-force but perfectly valid solution is to write *[tex \large (x+1)(x+2)(x+3)(x+4) + 1 = (x^2 + ax + b)^2] for some integers a and b and try to find a,b so that the equation is an identity. If we match the constant terms we see that b = 5. The coefficient of x^3 on the LHS is 10, so the coefficient of x^3 on the RHS must also be 10. If we were to take the x^3 coefficient on the RHS we would see that it is 2a, so 2a = 10 --> a = 5, and we have


*[tex \large (x+1)(x+2)(x+3)(x+4) + 1 = (x^2 + 5x + 5)^2]


which holds for all x.