Question

√a(a+1)(a+2)(a+3)+1∈N , oricare ar fi a∈N

Answer 1

$\rm \sqrt{a(a+1)(a+2)(a+3)+1} \in \mathbb N \\ \\ \sqrt{a(a+3)(a+1)(a+2)+1}= \\ \\ \sqrt{(a^2+3a)(a^2+2a+a+2)+1}= \sqrt{(a^2+3a)(a^2+3a+2)+1} \\ \\ Not\v am~a^2+3a=x: \\ \\ \sqrt{x(x+2)+1}=\sqrt{x^2+2x+1} \\ \\ x^2+2x+1=x^2+2\cdot x\cdot1+1^2=(x+1)^2 \\ \\ \sqrt{(x+1)^2} \geq 0 \Longrightarrow \boxed{\bold{\sqrt{(x+1)^2}~este~natural}}$