Question

n(n+1)(n+2)(n+3) + 1 sa fie pătrat perfect

Answer 1

 

$\displaystyle\bf\\n(n+1)(n+2)(n+3) + 1=\\\\=(n^2+n)(n^2+5n+6)+1=\\\\=n^4+n^3+5n^3+5n^2+6n^2+6n+1=\\\\= \boxed{\bf~n^4 + 6n^3 + 11n^2 + 6n + 1}\\\\.....................................................\\\\n^4 + 6n^3 + 11n^2 + 6n + 1 =\\\\= n^4 + 3n^3 + 3n^3 + n^2 + 9n^2 + n^2 + 3n + 3n + 1 =\\\\= (n^4 + 3n^3 + n^2) + (3n^3  + 9n^2 + 3n) + (n^2 + 3n + 1) =\\\\= n^2(n^2 + 3n + 1) + 3n(n^2  + 3n + 1) + (n^2 + 3n + 1) =\\\\= \boxed{\bf(n^2 + 3n + 1)(n^2 + 3n + 1)} = pp$

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$\displaystyle\bf\\\implies~n(n+1)(n+2)(n+3) + 1~~\text{este patrat perfect pentru oricare }~n\in N\\\\\\Exemple:\\\\n=0\implies~n(n+1)(n+2)(n+3) + 1 = 1=pp\\\\n=1\implies~n(n+1)(n+2)(n+3) + 1 = 25=5^2=pp\\\\n=2\implies~n(n+1)(n+2)(n+3) + 1 =121~=11^2=pp\\\\n=3\implies~n(n+1)(n+2)(n+3) + 1 = 361=19^2=pp\\\\n=4\implies~n(n+1)(n+2)(n+3) + 1 = 841=29^2=pp\\\\s.a.m.d.$