help with recurrence relation

I've been given this problem

Assuming that u(r) (ie. u subscript r) =a^(r) (a to the power r)

satisfies the recurrence relation u(r+2) = 5u(r+1) - 6u(r) (the
(r+2)th term = 5 times the term proceding - 6 times the term before
that)

show that a satisfies the equation a^2 - 5a + 6 = 0.


I've googled reccurence relations and the standard approach seems to be


u(r) = a^(r) implies that u^(r+1) = a^(r+1)

also u^(r+2) = a^(r+2)

Substitution into the recurrence relation gives

a^(r+2) = 5a^(r+1) - 6a^(r)

divide by a^(r) to get the quadratic.

http://en.wikipedia.org/wiki/Recurrence_relation

Clearly this approach is bogus.

u(r) = a^(r) in no way implies that a^(r+1) = u^(r+1).

In fact u(r+1) is clearly equal to 5( 5u(r) - 6u(r-1)) - 6 u(r)

that is

19u(r) - 30u(r-1)

for the given relationship. Indeed schoolboys use an approach similar
to wikipedia's to prove that one equals two.

The next part of the question is even worse.

"Hence find p and q such that u(r) = p^(r) and u(r) = q^(r) satisfy
the recurrence relation."

Clearly impossible for any non zero p and q.

I'm guessing the answer they want has something to do with the roots of
that quadratic, but I can't see how or why. I was tempted to dismiss
this question as being utter ***, but there are some intelligent
people knocking around that take this stuff seriously.

Can anyone please explain to me the error of my ways?

many thanks in advance.

.

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