Use induction to prove that $(3+ \sqrt 5)^n + (3-\sqrt 5)^n$ is always even.
Can someone please help me with this problem.
Use induction to prove that $(3+ \sqrt 5)^n + (3-\sqrt 5)^n$ is always even.
Can someone please help me with this problem.
Hint $\rm\,\ a^{n+2}\! + b^{n+2} = (a\!+\!b)\,(a^{n+1}\!+b^{n+1}) - ab\, (a^n\! + b^n).\ $ Here $\rm\:a\!+\!b = 6,\ ab = 4.$
Or: $ $ note that $\rm\:f(x) + f(-x) = 2\,(f_0 + f_2 x^2 + f_4 x^4 +\cdots\, + f_{2k}\, x^{2k}) =$ twice the even part of $\rm f(x).\:$ Therefore for $\rm\:f(x) = (3+x)^n$ and $\rm\:x = \sqrt{5},\:$ the RHS is an even integer by $\rm\,f_i\in \Bbb Z,\ x^{2i}\! = 5^i\in \Bbb Z.$
Prove by induction that $(3+ \sqrt 5)^n= x_n + y_n \sqrt 5$ for $x_n$ and $y_n$ integers because $x_0=1$ and $y_0=0$ and $x_{n+1}=3x_n+5y_n$, $y_{n+1}=x_n+3y_n$.
Prove by induction that $(3-\sqrt 5)^n=x_n - y_n \sqrt 5$, for the same $x_n$ and $y_n$ as above.
Conclude that $(3+ \sqrt 5)^n + (3-\sqrt 5)^n= 2x_n$, which is an even integer.
Hint: Binomial Theorem. ${}{}{}$