Question 89691
let x equal the first integer ... x(x+1)(x+2)(x+3)+1=x^4+6x^3+11x^2+6x+1


since this is a perfect square and the coefficients of the first and last terms are 1, the factors must be of the form x^2+bx+1


(x^2+bx+1)^2=x^4+2bx^3+(b^2+2)x^2+2bx+1 ... so b=3 ... x^2+3x+1=x(x+3)+1


integers that are separated by an odd amount (3) must be odd and even; so their product must be even ... adding 1 makes it odd


ALWAYS