O polinômio característico dessa recorrência é [tex3]x^2 =1996x+1997 [/tex3] , o qual tem raízes [tex3]-1[/tex3] e [tex3]1997 [/tex3] .
Então:
[tex3]\begin{cases}
1=x\cdot (-1)^2 + y\cdot (1997)^2 \\
3993= x\cdot (-1)^3+y\cdot (1997)^3
\end{cases}[/tex3]
Somando as 2:
[tex3]\frac{3994}{(1997)^3+(1997)^2} = y[/tex3]
[tex3]\frac{2}{(1997)^2+(1997)} = y[/tex3]
[tex3]\frac{1}{1997 \cdot 999} = y[/tex3]
[tex3]-\frac{998}{ 999} = x[/tex3]
Assim:
[tex3]x_{1997} = -\frac{998}{999} \cdot (-1)^{1997} + \frac{(1997)^{1997}}{1997 \cdot 999}[/tex3]
[tex3]x_{1997} = \frac{998+(1997)^{1996}}{ 999}[/tex3]
[tex3]x_{1997} = 1+\frac{(1997)^{1996}-1}{ 999}[/tex3]
[tex3]x_{1997} = 1+\frac{((1997)^{499})^{4}-1}{ 999}[/tex3]
[tex3]x_{1997} = 1+\frac{((1997)^{499}-1)((1997)^{499}+1)((1997)^{998}+1)}{ 999}[/tex3]
[tex3]x_{1997} = 1+\frac{((1997)^{499}-1)(1998)(1997^{498}-1997^{497}+1997^{496}-1997^{495}+....-1997+1)((1997)^{998}+1)}{ 999}[/tex3]
[tex3]x_{1997} = 1+2\cdot ((1997)^{499}-1)(1997^{498}-1997^{497}+1997^{496}-1997^{495}+....-1997+1)((1997)^{998}+1)[/tex3]
[tex3]x_{1997} \equiv 1+2\cdot ((-1)^{499}-1)((-1)^{498}-(-1)^{497}+(-1)^{496}-(-1)^{495}+....+1+1)((-1)^{998}+1)\mod(3)[/tex3]
[tex3]x_{1997} \equiv 1+2\cdot (-2)\cdot (499)\cdot (2)\mod(3)[/tex3]
[tex3]x_{1997} \equiv 1-8\mod(3)[/tex3]
[tex3]\boxed{x_{1997} \equiv 2\mod(3)}[/tex3]
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Problema 70
(Kosovo - 2013) Encontre todos os inteiros n tais que [tex3]\frac{n^2+n-27}{n-5}[/tex3] é um número inteiro.
2,4,6,8