Ans: A
Let $x$ be the smaller one of two consecutive integers, then the larger integer is $(x+1)$.
Let $x$ be the smaller one of two consecutive integers, then the larger integer is $(x+1)$.
Then the sum of the squares of the two integer
$\begin{array}{cl}
= & x^2 + (x+1)^2
\end{array}$
Three times the product of the two integers
$\begin{array}{cl}
= & 3 \times x \times (x+1) \\
= & 3x(x+1)
\end{array}$
Since the sum of the squares of the two integer is less than three times the product of the two integers by $1$, then we have
$\begin{array}{rcl}
x^2+(x+1)^2 & = & 3x(x+1) – 1
\end{array}$
