I've been given the following problem from a Putnam exam, and the solution I've generated seems far less complicated than the one I've seen. The problem motivation states:
Let $\{x_n\}$, $n\geq0$, be a sequence of nonzero real numbers such that ${x_n}^2-{x_{n-1}}{x_{n+1}}=1$ for $n=1,\ 2,\ 3,\ ...$. Prove there exists a real number $a$ such that ${x_{n+1}}=ax_n-x_{n-1}$ for all $n\geq1$.
My solution is essentially that $$a=\frac{x_{n+1}+x_{n-1}}{x_n}$$ therefore $$x_{n-1},\ x_n,\ x_{n+1}\ \epsilon\ \mathbb{R}^{\lt0}\cup\mathbb{R}^{\gt0}\ \implies a\ \epsilon\ \mathbb{R}$$
Not to be flip, but proving that such a number exists seems trivial. Is it implied that I should try to find an actual value for $a$? Have I actually proved that a real number must exist that fits the conditions of this sequence without having demonstrated a concrete value? I have this nagging feeling that I've left out something crucial, or that my solution is lacking in necessary rigor.
You must prove that $a$ is a constant NOT dependent on $n$.
Hint
In fact $$a={x_0^2+x_1^2-1\over x_0x_1}$$Substitute and conclude (it was rather a long time to me to prove it :D).