first of all we also need to assume that i've been trying for a couple of hours now to find a simpler proof for this problem but i haven't been able to do so! so, i'm just going to give

you my "not very simple" proof. we need a lemma:

Lemma: suppose and for some and integer then

Proof: by induction on if or there is nothing to prove. so suppose the claim is true for and where so we have for some

which gives us hence for some therefore and hence by the induction hypothesis and

so for some but then and so and we're done by the induction hypothesis.

now suppose for some we want to prove that either or obviously we may assume that both are non-

zero. let be the field of fractions of clearly in we have so we may assume that so there exists such that which we

will call it (1). let and where and so (1) becomes: call this (2).

the claim is that for all which will complete the solution because then and thus by (1): here's the proof of this claim:

from (2) we have: call this one (3). now, ignoring in (3), the rest of the relations can be written as where:

is an lower triangular matrix with on the main diagonal, an vector with the entries and an vector with the entries multiplying

from the left by gives us and so for all similarly if we ignore in (3) and write the remaining relations in terms of matrices, we will get

for all therefore for all by the lemma.