[tex]Notam:~b=a+1~;~c=a+2~si~d=a+3~(deoarece~numerele \\ \\ sunt~consecutive). \\ \\ abcd+1=a(a+1)(a+2)(a+3)+1= \\ \\ ~~~~~~~~~~~~= \big [a(a+3) \big] \big[(a+1)(a+2) \big]+1= \\ \\ ~~~~~~~~~~~~=(a^2+3a)(a^2+3a+2)+1. \\ \\ Notam~a^2+3a=n. \\ \\ abcd=n(n+2)+1=n^2+2n+1=(n+1)^2. [/tex]
[tex]Deoarece~abcd+1=(n+1)^2 \geq 0 \Rightarrow \sqrt{abcd+1}~are~sens \\ \\ pentru~orice~numere~intregi~consecutive~a,b,c,d. \\ \\ \sqrt{abcd+1}= \sqrt{(n+1)^2}=|n+1| \in Q.[/tex]