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In part of my research, the following problem has come up.

Consider the system of equations (in complex numbers)

$$z^b w^c = 1,\quad z^d w^e = 1.$$

I am interested in the solution set when we restrict both $z$ and $w$ to be $a^{\textrm{th}}$ roots of unity, for some positive integer $a$. Of course, one immediately sees that $(z, w) = (1, 1)$ is a solution.

What are some nice necessary and sufficient conditions on $a, b, c, d,$ and $e$ which guarantee that $(z, w) = (1, 1)$ is the ONLY solution?

To give an idea of the flavor of answer I'd be most happy with, one must have $\textrm{gcd}(a, b, d) = \textrm{gcd}(c, e) = 1$, because if $z$ is any $\textrm{gcd}(a, b, d)^{\textrm{th}}$ root of $1$ (which is neccesarily an $a^{\textrm{th}}$ root of $1$), then $(z, 1)$ is a solution to both equations.

It also turns out that $z$ and $w$ must both be $\textrm{gcd}(a, be - cd)^{\textrm{th}}$ roots of $1$.

I'd love to have an answer like "$\textrm{gcd}(a, be - cd) = \textrm{gcd}(a, b, d) = \textrm{gcd}(c, e) = 1$ is necessary and sufficient", with, perhaps, a few more estimates on gcd terms.

This problem can also been generalized (and I am interested in that case as well). Suppose you are given $3$ equations

$$z^a w^b = 1,\quad z^c w^d = 1,\quad z^e w^f = 1,$$

with $z$ and $w$ complex numbers of modulus $1$.

What are necessary and sufficient conditions on $a, b, c, d, e$ and $f$ which guarantee that the only simultaneous solution is $(z, w) = (1, 1)$?

The previous problem is a special case of this (which comes from setting $b = 0$. Clearly then, $z$ must be an ath root of unity. It turns that if $b$ is $0$, using the fact that $\textrm{gcd}(d, f) = 1$ one can show $w$ must also be an $a^{\textrm{th}}$ root of unity).

And please feel free to retag as appropriate!

Thank you in advance.

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1 Answer 1

up vote 5 down vote accepted

You are computing the nullspace of a $2 \times 2$ matrix over $\mathbb{Z}/a\mathbb{Z}$, so a necessary and sufficient condition is that the matrix in question is invertible, hence $\gcd(a, be-dc) = 1$ should be necessary and sufficient. In the second case you should compute the Smith normal form of your matrix over $\mathbb{Z}$. I believe a necessary and sufficient condition is that the Smith normal form is [[1 0][0 1][0 0]] (looks like matrices aren't working yet).

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Thank you, Qiaochu! I'm a bit embarrassed by the fact I couldn't figure this out! (Incidentally, the fact that it's a matrix over Z/aZ is equivalent to asking that (c,e)=1, so that neccesary condition I noticed is a bit hidden in your formulation). I'll be acknowledging you (and this question, once MO decides how to best reference itself) in my thesis! –  Jason DeVito Jul 30 '10 at 23:45
1  
I just wanted to add that I was able to prove that what you said about Smith normal form was exactly correct. As an added bonus, some google searching showed the product of the two "diagonal" terms (which, therefore, must be 1), is given as the gcd of all the determinants of 2x2 submatrices. So, in practice, I don't even have to compute the Smith normal form. Your answer turned out to be so helpful, I want to urge other people to vote it up! –  Jason DeVito Jul 31 '10 at 20:05
    
Thanks! Glad I could be of help. –  Qiaochu Yuan Aug 1 '10 at 5:29

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