Generating residues with $ a^n + b^n \mod p $

Question

Say there exist some non-zero distinct residues $a,b$ such that

$$ a^n + b^n \mod p $$

generates all nonzero residues for some $n$.

Does such a pair $a,b$ exist for every odd prime $p > 13$ ? Or for all sufficiently large prime $p$ ?

And if so, how many pairs exist as a function of $p$ ?

I consider $3$ cases :

1. $a$ is a primitive root, $b$ is not (or visa versa by symmetry)
2. Neither $a,b$ are primitive roots.
3. Both $a,b$ are primitive roots.

And I would like some insight about these cases.

edit

How about $a^n + b^n = a^n + (ac)^n = a^n(1 + c^n) \mod p$ ? Does that help ? Im not certain what and if " factoring " helps.

edit 2

To be clear it is not forbidden that $a^n + b^n = 0$ but it must generate all nonzero residues for some $n$. And this does not violate pigeonhole principles. See my " answer " that is more of a comment.

$n$ can go from $0$ to $p$ , after that it repeats by fermat's little rule.

edit 3

It is clear from the "answers and comments" that our best bet for having a solution seems to be the primes $p>13$ such that $(p-1)/2$ is also a prime. These primes are called Sophie Germain primes.

( for number theory fans, I noticed something weird about those primes , see : Why does this ratio $5$ occur relating prime twins and sophie germain primes?, see also https://en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes
)

edit 4

ok based on some mistakes, edit 3 was exgaggerated.

$p = 4 m + 3$ are already very good candidates.

So perhaps we should look at those in particular

$3,7,11,19,23,...$

Answer 1

General conditions for a solution

Let $b=ac$ where $c$ has order $q$ and so

$$a^n+b^n=a^n(1+c^n).$$

It is convenient to consider the necessary conditions in group theoretic terms, where $G$ is the group of non-zero residues modulo $p$.

Let $H=<a^q>$ and let $Q=\{a^n(1+c^n)|0\le n\le q-1\}$. Then $|H|\le \frac{p-1}{q}$ and $|Q|\le q$.

We require $G\subseteq HQ$ and so we require the above bounds to be attained. Therefore

• $c^n\ne -1$ and so $q$ is odd
• $H$ is the subgroup of all $q$th powers
• Each element of $Q$ is in a different coset of $H$

$G/H$ is a group of odd order and so the product of its elements is the identity. Therefore $$\prod_{0}^{q-1}a^n\prod_{0}^{q-1}(1+c^n)\in H.$$ The product of the $a^n$ is $a^\frac{q(q-1)}{2}\in H$ and the $1+c^n$ satisfy $(x-1)^q-1=0$ and so have product $-2$. Therefore

• $2\in H$

The case $q=3$.

As found by @The Phoenix, small solutions appear to be of this type and so we will do a full analysis of this case and show that there are precisely $\phi (p-1)$ pairs of solutions for $a$ and $b$.

We first note that $p \equiv 1 \bmod{3}$ and that $2$ is a cubic residue. Therefore we are concerned with primes $p$ of the form $x^2+27y^2$, the smallest such prime being $31$.

$1+c+c^2=0$ and so $Q=\{2,a(1+c),a^2(1+c^2)\}=\{2,-ac^2,-a^2c\}$. The coset condition therefore reduces to simply $\frac{a}{c}\notin H$.

THE CASE $p=9k+1$

In this case $c$ is a cube and so $a$ is not a cube. $a^3$ generates $H$ and so $a$ is a primitive root. Then $c=a^{3k}$ or $a^{6k}$.

There are $\phi (p-1)$ primitive roots and each $a$ is paired with two $b$s given by $b=a^{1+3k}$ and $b=a^{1+6k}$. Note that if $b=a^{1+6k}$ then $a=b^{1+3k}$.

• $a$ and $b$ are one of $\phi (p-1)$ pairs of primitive roots.

THE CASE $p=9k+4$

First suppose $a$ is a primitive root then $c=a^{3k+1}$ or $a^{6k+2}$. Since $\frac{a}{c}\notin H$ we must have the latter possibility.

There is also the possibility that $a^3$ generates $H$ without $a$ being a primitive root. In this case $b$ is a primitive root and we can simply interchange the nomenclature $a$ and $b$.

• $a$ is one of $\phi (p-1)$ primitive roots and $b=a^{6k+3}$.

THE CASE $p=9k+7$

As in the previous case we can suppose $a$ is a primitive root and then $c=a^{3k+2}$.

• $a$ is one of $\phi (p-1)$ primitive roots and $b=a^{3k+3}$.

ILLUSTRATIVE EXAMPLE $p=31$

The $\phi (p-1)=8$ primitive roots are $3,11,12,13,17,21,22,24.$

$k=3$ and so each $b=a^{21}$.

The $(a,b)$ pairs are $$(3,15),(13,15),(17,23),(22,23),(11,27),(24,27),(12,29),(21,29).$$

The case $q=5$.

A similar analysis can be carried out for larger $q$. In the next simplest case, of $q=5$, two conditions are required:

• $2\in H$
• $\sqrt 5-1\in H$

To illustrate the connection with the $p=9k+1$ case analysed above let us consider primes of the form $p=25k+1$. Those which satisfy the condition for $2$ are $$151,251,3251,3301,8501,...$$

Of these first $5$ primes, only $3251$ satisfies the condition for $\sqrt 5-1$.

The solutions themselves are constructed in exactly the same way as those for $q=3$. The $\phi (p-1)$ primitive roots are split up into appropriate sets of $5$: $$\{ 6,190,1682,2293,2331\},\{ 10,634,636,2484,2738\}, \{ 37,88,1678,1703,2996\}, ...$$ and then solutions are given by choosing $a$ and $b$ from the same set. The number of pairs is then ${5 \choose 2} \frac{\phi (p-1)}{5}$ compared to ${3 \choose 2} \frac{\phi (p-1)}{3}$ for $q=3$.

Answer 2

Here are some results:

Claim One: If $2$ is a primitive root modulo $p$, then there do not exist $a$ and $b \ne a$ such that $a^n + b^n$ represents all non-zero residue classes modulo $p$. More precisely, if there exists such $a$ and $b$ then there is an odd prime $q$ such that $p \equiv 1 \bmod q$ and in addition $2 \equiv u^q \bmod p$ for some $u \bmod p$.

Artin's Conjecture (proved under GRH) says that $2$ is a primitive root for a positive proportion of primes $p$ (and certainly infinitely often). Hence this strongly suggests (and proves, assuming the General Riemann Hypothesis) the answer to both questions is no, there will not exist such $a$ and $b$ for large enough $p$. As a partial converse, however, we have:

Claim Two: If $p \equiv 1 \bmod 3$ and $2$ is a cube modulo $p$, then there does exist $a$ and $b \ne a$ such that $a^n + b^n$ represents all residue classes modulo $p$.

The relative density of primes $p$ satisfying this last condition is $1/6$ (that is, asymptotically a sixth of the primes satisfy this condition). These are also primes which can be written in the form $p = x^2 + 27 y^2$. So the summary seems to be that such $a$ and $b$ exist for a positive proportion of primes $p$ and also do not exist for a positive proportion of primes $p$.

Computationally, the primes $p < 500$ for which $a$ and $b$ exist are

$$31,43,109,127,157,223,229,277, 283, 307, 397, 433, 439, 457, 499, \ldots $$

and these are exactly the primes for which $p \equiv 1 \bmod 3$ and $2$ is a cube. I tried for some time to prove that this condition is necessary as well as sufficient but failed. But that is because it turns out not to be a necessary condition, if $p = 3251$, then $2 \equiv 40^5 \bmod 3251$ and $3251 \equiv 1 \bmod 5$. Note that $p \not\equiv 1 \bmod 3$. But

$$6^n + 190^n \bmod 3251$$

represents all non-zero congruence classes. So ultimately the density $\varrho$ of primes $p$ with this property (assuming GRH) is somewhere (certainly strictly) in the range:

$$0.16666 \ldots = 1/6 \le \varrho \le 1 - \prod_{p > 2} \left(1 - \frac{1}{p(p-1)}\right) = 0.25208 \ldots $$

Proof of Claim One: Here are the arguments. By Fermat's Little Theorem, $a^n + b^n \bmod p$ has period (dividing) $p-1$, so if it generates all non-zero residues it generates exactly those residues exactly once. Write $b = ac$. Since $a^n (1 + c^n)$ can never equal $0$, it follows that $c^n$ can never equal $-1$, which means that $c$ has odd order. If follows that $c$ is a quadratic residue. Hence $a$ is not a quadratic residue since otherwise the period of $a^n + b^n$ would divide $p-1$. Thus $a^{(p-1)/2} \equiv -1 \bmod p$. By Wilson's theorem we have

$$-1 \equiv (p-1)! \equiv \prod_{i=1}^{p-1} i,$$

and thus $$-1 \equiv \prod_{n=0}^{p-2} (a^n + b^n) = \prod_{n=0}^{p-2} a^n \prod_{n=0}^{p-2} (1 + c^n) = (a^{(p-1)/2})^{p-2} \prod_{n=0}^{p-2} (1 + c^n).$$

Here the first equality comes from the fact that as $n$ ranges over this set the numbers $a^n+b^n$ are just the integers from $1$ to $p-1$ modulo $p$ albeit in a different order. Since we have already shown that $a$ is not a quadratic residue, the power of $a$ is equal to $-1$, so we deduce that

$$1 \equiv \prod_{n=0}^{p-2} (1 + c^n).$$

Now suppose that the order of $c$ is $d$ (which we know is odd), with $(p-1) = de$. Then the product above repeats $e$ times, and we get

$$1 \equiv \left( \prod_{n=0}^{d-1} (1 + c^n) \right)^e.$$

Let's evaluate the interior product. Since $c$ has order exactly $d$, we have a factorization

$$X^d - 1 = \prod_{n=0}^{d-1} (X - c^n).$$

Thus we have (using that $d$ is odd several times)

$$\prod_{i=0}^{d-1} (1 + c^i) = (-1)^d \prod_{i=0}^{d-1} (-1 - c^i) = - \lim_{X \rightarrow -1} \prod_{i=0}^{d-1} (X - c^i)$$ $$ = - \lim_{X \rightarrow -1} (X^d - 1) = - ((-1)^d - 1) = 2.$$

So putting this together, we deduce that

$$2^e \equiv 1 \bmod p$$

where $c$ has order $d$ and $p-1 = de$. It follows that $2$ is a $d$th power modulo $p$. Remember that $d > 1$ is some odd divisor of $p-1$, so there is certainly an odd prime $q > 1$ dividing $d$ and $2$ is a perfect $q$th power and $p \equiv 1 \bmod p$. In particular, this proves that $2$ is not a primitive root modulo $p$.

Proof of Claim Two: Now let's move on to the second claim, so assume that $p \equiv 1 \bmod 3$ and $2$ is a cubic residue modulo $p$. Note that these are primes which split completely in $\mathbf{Q}(\sqrt[3]{2},\sqrt{-3})$, so the $1/6$ density claim follows from the Cebotarev density theorem.

Some preliminaries. Suppose that $\omega^3 = 1$ but $\omega \ne 1$. Then $\omega^2 + \omega + 1 = 0$, and $\omega^{2n} + \omega^n + 1 = 0$ for $(3,n) = 1$. In particular, if $(3,n) = 1$, then

$$1 + \omega^n = - \omega^{2n}, \quad \frac{1}{1 + \omega^n} = - \omega^n.$$

Now let us begin the construction. Let $\varepsilon$ be a primitive root modulo $p$. Let $\omega$ be an element which satisfies $\omega^3 = 1$ but $\omega \ne 1$. This exists because $p \equiv 1 \bmod 3$. We choose $\omega$ to satisfy one further property, namely that

$$\varepsilon/\omega \ \text{is not a cube modulo $p$}.$$

We can always achieve this; if $p \equiv 1 \bmod 9$ then $\omega$ is a cube so this is automatic. If $p - 1$ is exactly divisible by $3$, then $\omega$ is not a cube, so either $\varepsilon/\omega$ or $\varepsilon/\omega^2$ is not a cube, and in the latter case simply replace $\omega$ by $\omega^2$.

Now let $a = \varepsilon$ and $b = \varepsilon \omega$. Note that $a^n + b^n = a^n (1 + \omega^n)$, since $1 + \omega^n \in \{-\omega, - \omega^2, 2\}$ this sum never vanishes. So it suffices to show that $a^n + b^n$ and $a^m + b^m$ are distinct if $0 \le n,m < p-1$ and $n \ne m$. If $a^n + b^n = a^m + b^m$, it follows that

$$\varepsilon^n (1 + \omega^n) = \varepsilon^m (1 + \omega^m), \quad \text{or} \quad \varepsilon^{n-m} = \frac{1 + \omega^m}{1 + \omega^n}.$$

We consider various cases depending on $m$ and $n$ modulo $3$. Note there is symmetry in $m$ and $n$ so we need not consider all cases.

1. Suppose that $m \equiv n \bmod 3$. Then $1+\omega^n = 1 + \omega^m$, and we deduce that $\varepsilon^{m-n} = 1$ which implies that $m=n$ because $\varepsilon$ is a primitive element.

2. Suppose that $3|m$ but $(3,n)=1$. Then

$$\varepsilon^{n-m} = - 2 \omega^n.$$

Since $-1$ is always a cube and $2$ is a cube by the assumption on $p$, and $\varepsilon^m$ is certainly a cube because $3|m$, we deduce that $\varepsilon^n/\omega^n$ is a cube, which since $(3,n)=1$ implies that $\varepsilon/\omega$ is a cube, contradicting the choice of $\omega$.

3. Suppose that $m \equiv 1 \bmod 3$ and $n \equiv 2 \bmod 3$. Then

$$\varepsilon^{n-m} = \frac{1 + \omega}{1 + \omega^2} = \omega,$$

and now since $\varepsilon^{n-m-1}$ is a cube, we again deduce that $\varepsilon/\omega$ is a cube, a contradiction.

If you assume that $p \equiv 1 \bmod 5$ and $(2/p)_5 = 1$, you can try something similar, taking $a$ to be a primitive root and $b=ac$ with $c = \zeta$ some carefully chosen $5$th root of unity. This generally doesn't work, because now the ratios

$$\frac{1 + \zeta^i}{1 + \zeta^j}$$

don't simplify and you don't have much control of their $5$th power residue symbols. But you can impose some conditions on the images of these elements in $\mathbf{F}^{\times}_p/\mathbf{F}^{\times 5}_p$ for things to work. And it does work for $p=3251$, the example $a=6$ and $b =190$ has $c = b/a = 2199 \bmod p$ with $c^5 \equiv 1 \bmod p$. For more general $p$, you can impose some conditions on the $5$th power residue symbols of the ratios $(1 + \zeta^i)/(1 + \zeta^j)$ for this to work, and this example shows those conditions will sometimes (but not always) be satisfied.

Added, May 30: You ask about the case $a^n + 2 b^n$. That is quite similar, although there are more primes $p$ for which this works. Here are some more comments.

If $c = b/a = - 1$, you are in good shape if either $3$ is a quadratic residue or $3$ is not a quadratic residue but $-1$ is also not a quadratic residue. So this covers the case

$$p \equiv 1,7,11 \bmod 12,$$

but leaves open $p \equiv 5 \mod 12$. But again in this case one does not expect to cover all cases. As before,

$$-1 = \prod (a^n + 2 b^n) = a^{(p-1)/2} \prod_{n=0}^{p-2} (1 + c^n),$$

and assuming that $c$ has order exactly $d$ with $de = p - 1$, you can show as above that:

$$-1 = a^{(p-1)/2} \left(1 - (-2)^d\right)^e \bmod p$$

Now suppose that $p = 4q+1$ is a prime such that $q \equiv 7 \bmod 30$, so $p \equiv 29 \bmod 120$ and in particular $p \equiv 5 \mod 12$ and $p \equiv 5 \bmod 8$ and $p \equiv -1 \bmod 5$. Standard conjectures say that there should be infinitely many primes of this form where $q$ is also prime; the first example is $p = 29$ and $q = 7$.

Claim: for $p = 4q+1$ prime with $q \equiv 7 \bmod 30$ prime, there are no $a$ and $b$ such that $a^n + 2 b^n$ cover all non-zero congruence classes. Standard conjectures imply that there are infinitely many such primes.

Proof: For such primes, the possible order of $c$ is $1$, $2$, $4$, $q$, $2q$, or $4q$. Let's consider each case in turn:

1. If $c$ has order $1$ then $c=1$ (which is not allowed).
2. If $c$ has order $2$ then $c=-1$, and this was considered above, and doesn't work when $p \equiv 5 \mod 12$.
3. If $c$ has order $4$, then $d = 4$, so $1 - (-2)^d = -15$. We get $e = (p-1)/4$, and so

$$- 1 = \pm 1 \cdot 15^{(p-1)/4} \bmod p,$$

or squaring both sides

$$1 = 15^{(p-1)/2} \bmod p.$$

But the congruences on $p$ imply that $15$ is not a quadratic residue modulo $p$, so $15^{(p-1)/2} \equiv -1 \bmod p$ and this is a contradiction.

4. Suppose that $c$ has order $q$. Since $q \equiv 5 \mod 8$, $-2$ is not a quadratic residue. That implies that if $d = (p-1)/4$ and $e=4$ then $2^d$ is a $4$th root of unity $i$. We deduce that

$$-1 = \pm 1 \cdot (1 + i)^4 \equiv \pm 4 \bmod p,$$

which is not possible.

5. Suppose that $c$ has order $q$. Since $q \equiv 5 \mod 8$, $-2$ is not a quadratic residue. That implies that if $d = (p-1)/2$ and $e=2$ then $2^d$ is a $2$th root of unity $-1$. We deduce that

$$-1 = \pm 1 \cdot (1 -1)^2 \equiv \pm 4 \bmod p,$$

which is not possible again.

6. Suppose that $c$ is a primitive root; then some power of $c$ is $-2$ and the expression vanishes.

Since this covers all cases, this completes the proof!

Answer 3

Just a comment based upon user1172706 answer :

It SEEMS in general if $(p-1)/2$ factors alot, then the pigeonhole principle together with fermat's little principle forbids us to generate every residue because we generated too many zeros.

In particular the number of times we can factor radical roots of minus unity out ($(x^s + 1)$) is equal to the number of times we can divide by 2 because we use $(x^2 - a^2) = (x-a)(x+a)$.

So in conclusion, heuristically if $p-1 = 2^K L$ for large $K$, then the probability for a solution is lowest ... I see that now from correctly factoring ...

Or equivalently for a prime $p$ :

$$p-1 = 2(2m + 1)$$

$$p-1 = 4 m + 2$$

$p = 4m + 3$

has the best odds of having a successful pair $a,b$.

Funny, I noticed $x^2 + y^2$ are primes of the form $4n+1$, not sure if it relates or not. ( $x^2 + y^2$ is the case $n=2$ )

We can also say that $a^n$ must generate at least $g$ distinct integers and $b^n$ must generate at least $h$ distinct integers , where we have $p-1 \leq gh$. (order is the right terminology for $g,h$ I believe)

Answer 4

Of course, a choice of $a,b$ exist for every prime. Simply choose $a$ to be a primitive root, and $b$ to be 1.

Let $a$ have order $\alpha$ and $b$ have order $\beta$ in the multiplicative group $\mathbb Z_p^*$, meaning $\alpha$ and $\beta$ are minimal and still solving the equation $a^\alpha=b^\beta=1\pmod p$.

The number of distinct elements of the form $a^n+b^n$ cannot exceed $LCM(\alpha, \beta)$. So, if $LCM(\alpha, \beta) < p$, you will never cover every element. For example, $p=17$, $a=4$, $b=2$. In this case, $\alpha = 4$ and $\beta=8$, but $LCM(4,8)=8 < 17$, so the equation $\ell = 4^n+2^n \pmod {17}$ cannot be solved for every $\ell \in \mathbb Z_{17}^*$.

Fermat's theorem tells us that $a^{p-1}=b^{p-1}=1 \pmod p$, so that each of $\alpha$ and $\beta$ divide $p-1$. So the factorization of $p-1$ will play a crucial role here. Take again $p=17$. Indeed, due to the factorization of $p-1=17-1=2^4$, we know that regardless of choice of $a, b$, that the LCM of the orders is equal to the max of the orders. And therefore, we know that all of the elements can be covered if and only if at least one of $a$ and $b$ are a primitive root.

The short answer is that it depends on your number $p$, $p-1$, and how well the particular choice adds up to produce distinct numbers modulo $p$. That is, how close you actually get to LCM distinct sums.