A root of a product plus one

Evaluate $\sqrt{120 \cdot 121 \cdot 122 \cdot 123 + 1}$ .

Show hints (2)+

Pair the factors: $n(n+3)$ and $(n+1)(n+2)$ differ by $2$ , straddling $m=n^2+3n+1$ .
So the product $+1$ is $(m-1)(m+1)+1=m^2$ . Compute $m$ for $n=120$ .

Answer

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14761

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