# Python’s “map” method and permutations of lists

Posted by Jason Polak on 18. February 2018 · Write a comment · Categories: computer-science · Tags: , ,

Let’s look at Python’s map function. What does this function do? Here is the syntax:

It takes a function called function and applies it to each element of iterable, returning the result as an iterable. For example:

The output of this program is:

More »

# Polynomial over finite field: permutation polynomial?

Posted by Jason Polak on 18. February 2018 · 2 comments · Categories: computer-science · Tags: , ,

Let’s assume you have a polynomial over a finite field $\F_q$, defined in Sage. How can you tell whether it’s a permutation polynomial? That is, when is the corresponding function $\F_q\to\F_q$ bijective?

This is how you might have a polynomial over $\F_q$ defined in Sage:

Here, the variable $x$ refers the element $x$ in the isomorphism $\F_q \cong \F_p[x]/\alpha(x)$ and $t$ is the variable in the polynomial ring $\F_q[t]$. Is $f$ a permutation polynomial? That of course depends on what $q$ is.
More »

# Same multiplicative order modulo p and p^2

In the abelian group $\Z/n$, the order of $m\in \Z/n$ can be calculated via the formula $n/{\rm gcd}(m,n)$. This number is just the smallest number you have to multiply $m$ by in order to get a multiple of $n$. So when $n = p$ is a prime, every we see that every nonzero element of $\Z/p$ has order $p$.

However, every nonzero element of $\Z/p$ is prime to $p$, and the nonzero elements of $\Z/p$ form the abelian multiplicative group $\Z/p^\times$ of invertible elements. Then the order of an element $m\in \Z/p^\times$ is a little more tricky to figure out, but we know by Fermat’s Little Theorem that it will be a factor of $p-1$.

Example. Consider the prime $p=11$. Then we know that $3^{10} = 1$ by Fermat’s Little Theorem. However, $3^5 = 1$ too. In fact the order of $~3$ is $~5$.

This example has a curious property that you may not know about. It is true that $3^5 = 1$ in $\Z/11$. But did you know that $3^5 = 1$ in $\Z/121$? What?! This leads to a natural question:

Question. Which primes $p$ have the property that there exists an element $m\in \Z/p^\times, m\not=1$ such that the order of $m$ in $\Z/p^\times$ is the same as the order of $m$ in ${\Z/p^2}^\times$?

Not all primes have this property. Consider $p = 5$. Then $3^4 = 1$ in $\Z/5$. However, $3^4 = 6$ in $\Z/25$. In fact, none of the elements $2,3,4$ have the same order in both $\Z/5^\times$ and $\Z/25^\times$. Here are some primes that do have such elements:

11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, …

So, there are quite a few of them and maybe there are infinitely many. One can go further with this question: what about primes for which there are non-trivial elements in $\Z/p^\times$ with the same order in $\Z/p^\times, {\Z/p^2}^\times,$ and ${\Z/p^3}^\times$? There are fewer of these. But they exist. For example, the identity
$$68^{112} = 1$$
Holds in $\Z/113^k$ for $k=1,2,3$ (but not for $k=4$). This is the only example I know of and maybe it is the only possible example? But I probably just haven’t looked far enough. I wrote the original program to find it in Python but I plan to rewrite it in C with the gmp library to expand the search.