Question

Determinați valorile întregi ale lui n pentru care radical din n^2-4n+13 e N.

Answer 1

Răspuns:

Explicație pas cu pas:

O zi frumoasă!

Answer 2

$\displaystyle\it\\\sqrt{n^2-4n+13} \in\mathbb{N} \Leftrightarrow n^2-4n+13~este~patrat~perfect.\\fie~m\in\mathbb{N}~a.i.~n^2-4n+13=m^2 \Leftrightarrow n^2-4n+4+9=m^2 \Leftrightarrow\\(n-2)^2+9=m^2 \Leftrightarrow 9=m^2-(n-2)^2 \Leftrightarrow 9=(m-n+2)(m+n-2).\\9=9\cdot1=1\cdot9=(-9)\cdot(-1)=(-1)\cdot(-9)=3\cdot3=(-3)\cdot(-3).$

$\displaystyle\it\\daca~9=9\cdot1 \implies \left \{ {{m-n+2=9} \atop {m+n-2=1}} \right. \Leftrightarrow \left \{ {{m-n=7} \atop {m+n=3}} \right. \implies \\2m=10 \implies \boxed{\it m=5 \implies n=-2}~.\\daca~9=1\cdot9 \implies \left \{ {{m-n+2=1} \atop {m+n-2=9}} \Leftrightarrow \left { {{m-n=-1} \atop {m+n=11}} \right \implies\\\\2m=10 \implies \boxed{\it m=5 \implies n=6}~.$

$\displaystyle\it\\daca~9=(-9)(-1) \implies \left \{ {{m-n+2=-9} \atop {m+n-2=-1}} \right. \Leftrightarrow \left \{ {{m-n=-11} \atop {m+n=1}} \right. \implies\\2m=-10 \implies \boxed{\it m=-5 \implies n=6}~.\\daca~9=(-1)(-9) \implies \left \{ {{m-n+2=-1} \atop {m+n-2=-9}} \right. \Leftrightarrow \left \{ {{m-n=-3} \atop {m+n=-7}} \right. \implies\\2m=-10 \implies \boxed{\it m=-5 \implies n=-2}~.$

$\displaystyle\it\\daca~9=3\cdot3 \implies \left \{ {{m-n+2=3} \atop {m+n-2=3}} \right. \Leftrightarrow \left \{ {{m-n=1} \atop {m+n=5}} \right. \implies\\2m=6 \implies \boxed{\it m=3 \implies n=2}.\\daca~9=(-3)(-3) \implies \left \{ {{m-n+2=-3} \atop {m+n-2=-3}} \right. \implies \left \{ {{m-n=-5} \atop {m+n=-1}} \right. \implies\\2m=-6 \implies \boxed{\it m=-3 \implies n=2}~.$