Lifting The Exponent

The "lifting the exponent" (LTE) lemma is a useful one about the largest power of a prime dividing a difference or sum of $n^\text{th}$ powers. Here are some sample problems whose solutions use the lemma:

1. Let $n$ be a square-free integer. Show that there is no pair of coprime positive integers $(x,y)$ such that $(x+y)^3 \big| (x^n+y^n).$
2. Show that $2$ is a primitive root mod $3^k$ for all positive $k$.
3. Find all solutions in positive integers to $3^n = x^k + y^k$, where $\gcd(x,y) = 1$, $k \ge 2$.
4. Suppose $a$ and $b$ are positive real numbers such that $a-b, a^2-b^2, a^3-b^3, \ldots$ are all positive integers. Show that $a$ and $b$ must be positive integers.

Let $p$ be a prime and $n$ a nonzero integer. Then we define $v_p(n)$ to be the exponent of $p$ in the prime factorization of $n$. That is,

$v_p(n) = k \Longleftrightarrow p^k\big|n \ \text{and} \ p^{k+1} \nmid n.$

Let $p$ be a prime, $x$ and $y$ integers, $n$ a positive integer, and suppose that $p|(x-y)$ but $p\nmid x$ and $p\nmid y$. Then

(1) if $p$ is odd, $v_p(x^n-y^n) = v_p(x-y) + v_p(n);$ (2) for $p = 2$ and even $n$, $v_2(x^n-y^n) = v_2(x-y) + v_2(n) + v_2(x+y) - 1.$

Notice that if $n$ is odd, we can substitute $-y$ for $y$ in (1) to obtain

$v_p(x^n+y^n) = v_p(x+y)+v_p(n)$

if $p|(x+y)$.

Use the LTE lemma to find the largest power of $3$ dividing $5^{18}-2^{18}$.

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By LTE,

$v_3\big(5^{18}-2^{18}\big) = v_3(5-2)+v_3(18) = 1+2 = 3.$

So the answer is $3^3$.

Without LTE, the problem can be solved by factoring

$\begin{aligned} 5^{18}-2^{18}&=\big(5^9-2^9\big)\big(5^9+2^9\big) \\ &= \big(5^3-2^3\big)\big(2^6+2^35^3+5^6\big)\big(5^9+2^9\big) \\ &= (5-2)\big(2^2+(2)(5)+5^2\big)\big(2^6+2^35^3+5^6\big)\big(5^9+2^9\big). \end{aligned}$

The first factor has one $3$ and the fourth factor has no $3$s, and some careful mod-$9$ analysis shows that the second and third factors are divisible by $3$ but not $9$, so the total number of factors of $3$ is $3$. This is quite a bit more complicated (but note that it also indicates how an inductive proof of LTE might proceed). $_\square$

This lemma gives a practical way to solve many problems involving the largest power of a prime that divides certain expressions. In particular, the solutions to the problems in the introduction all use LTE in an essential way.

As a warmup, here is a typical Diophantine equation that can be tackled using the LTE Lemma.

Let $n$ be a square-free integer. Show that there is no pair of coprime positive integers $(x,y)$ such that

$(x+y)^3 \Big| (x^n+y^n).$

Assume $(x+y)^3 \Big|(x^n+y^n)$ with $\gcd (x,y)=1$. We will derive a contradiction.

First, suppose $n$ is even. If there is an odd prime $p |(x+y)$, then $x^n+y^n \equiv x^n +(-x)^n \equiv 2x^n$ mod $p$, so $p|x$, so $p|y$, contradiction. Since $x$ and $y$ are positive, the only possible way that there is no odd prime $p$ dividing $x+y$ is if $x+y$ is a power of $2$. In this case, $x$ and $y$ are both odd since they are coprime, so since $n$ is even $x^n$ and $y^n$ are both $1$ mod $8$, it follows that $v_2(x^n+y^n) = 1$, but $v_2\big((x+y)^3\big) \ge 3$, so again we get a contradiction.

Now suppose $n$ is odd. If there is an odd prime $p|(x+y)$, then $3 v_p(x+y) \le v_p(x^n+y^n)$. Then LTE gives

$3 v_p(x+y) \le v_p(x^n+y^n) = v_p(x+y) + v_p(n) \le v_p(x+y)+1$

since $n$ is square-free. Simplifying, we get $v_p(x+y) \le \frac12$, which is impossible since $p|(x+y)$.

As above, the only remaining case is when $x+y$ is a power of $2$. But in this case, $v_2(x^n+y^n) = v_2(x+y)$ if $n$ is odd, because $x$ and $y$ are odd and the expansion of

$\frac{x^n+y^n}{x+y} = x^{n-1} -x^{n-2}y + \cdots - xy^{n-2}+y^{n-1}$

has an odd number of terms, all of which are odd. So it is impossible for $(x+y)^3$ to divide $x^n+y^n$, since the power of $2$ on the left exceeds the power of $2$ on the right. $_\square$

Show that $2$ is a primitive root mod $3^k$ for all positive $k$.

We seek the smallest positive $n$ such that $2^n \equiv 1$ mod $3^k$. If $3^k \big|(2^n-1)$, first note that $2^n \equiv 1$ mod $3$, so $n$ is even. Write $n = 2m$. So $3^k \big|(4^m-1)$. Now LTE applies:

$k \le v_3(4^m-1^m) = v_3(4-1)+v_3(m) = 1 +v_3(m).$

So $v_3(m) \ge k-1$. The smallest possible such $m$ is $3^{k-1}$, so the smallest possible $n$ is $2\cdot 3^{k-1} = \phi(3^k)$, where $\phi$ is Euler's totient function. This proves the claim. $_\square$

Here is a similar problem to try:

Find all solutions in positive integers to $3^n = x^k + y^k$, where $\gcd(x,y) = 1$, $k \ge 2$.

If one of $x$ or $y$ is divisible by $3$, then they both are, which is a contradiction. So neither is. If $k$ is even, then $x^k$ and $y^k$ are both $1$ mod $3$, so $x^k+y^k$ is not divisible by any power of $3$.

Now suppose $k$ is odd. If $n=0,$ then $x^k+y^k = 1$, and there are no solutions to this in positive integers, so we can exclude this case. Since $n \ge 1$, $3|(x+y)$. Apply LTE:

$n = v_3(x^k+y^k) = v_3(x+y) + v_3(k).$

Then

$x^k + y^k = 3^n = 3^{v_3(x+y)}3^{v_3(k)} = (x+y)k.$

The point is that the left side is usually much bigger than the right side, so the result will follow from some routine inequalities.

Suppose $x > y$ without loss of generality. Then dividing through by $x+y$ gives

$\begin{aligned} x^{k-1} - x^{k-2}y + \cdots - xy^{k-2} + y^{k-1} &= k \\ (x-y)\big(x^{k-2} +x^{k-4}y^2+ \cdots + xy^{k-3}\big) + y^{k-1} &=k. \end{aligned}$

The left side is $\ge x^{k-2}$, so $x^{k-2} \le k$. So $\ln(x) \le \frac{\ln(k)}{k-2}$. Recall $k \ge 3, x \ge 2$. By calculus, the right side is decreasing for $k \ge 3$, so $\ln(x) \le \ln(3)$, so $x \le 3$. We already ruled out $3|x$, so $x= 2$ and hence $y = 1$.

In that case $2^{k-2} > k$ already unless $k=3,4$, but $k$ is odd so $k = 3$. It's easy to check that $x=2,y=1,k=3$ is a solution, as is $x=1,y=2,k=3$ if we relax the $x>y$ assumption, and the above analysis has shown that these are the only ones. $_\square$

Suppose $a$ and $b$ are positive real numbers such that $a-b, a^2-b^2, a^3-b^3, \ldots$ are all positive integers. Show that $a$ and $b$ must be positive integers.

Since $a-b\in\mathbb Z$ and $a^2-b^2 \in \mathbb Z$, $a+b= \frac{a^2-b^2}{a-b} \in \mathbb Q$. But then $a = \frac12(a-b) + \frac12(a+b)$ and $b = \frac12(a+b)-\frac12(a-b)$ are rational numbers as well.

Now, write $a = \frac{x}{z}$ and $b = \frac{y}{z}$ as quotients of positive integers, with a common denominator. Choose $z$ as small as possible. Then the conditions of the problem imply that

$z^n \big| (x^n-y^n) \ \text{for all} \ n.$

Suppose $p$ is a prime dividing $z$. Note $z|(x-y)$ so $p|(x-y)$. If $p|x,$ then $p|y$ as well, but that violates the choice of $z$: we could write $a = \frac{\hspace{1.5mm}\frac xp\hspace{1.5mm}}{\hspace{1.5mm}\frac zp\hspace{1.5mm}}, b = \frac{\hspace{1.5mm}\frac yp\hspace{1.5mm}}{\hspace{1.5mm}\frac zp\hspace{1.5mm}}$ to get a smaller common denominator.

So $p \nmid x,y$ and we are set up to apply LTE. If $p$ is odd, then

$n \le v_p(z^n) = v_p(x^n-y^n) = v_p(x-y) + v_p(n).$

Taking $p$ to both sides gives

$\begin{aligned} p^n &\le (x-y)n \\ \frac{p^n}{n} &\le (x-y) \end{aligned}$

but this is impossible since the left side goes to infinity as $n \to \infty$, and the right side is a constant independent of $n$.

If $p = 2,$ we get

$n \le v_2(z^n) = v_2(x^n-y^n) = v_2(x-y)+v_2(n)+v_2(x+y)-1,$

so

$\frac{2^{n+1}}{n} \le (x-y)(x+y)$

and we get a similar contradiction.

The conclusion is that there is no prime $p$ dividing $z$. So $z=1$ and $a$ and $b$ are both positive integers. $_\square$

Here is an outline of the proof for odd primes $p$; the proof for $p = 2$ is similar. Suppose $p | (x-y)$, $p \nmid x, p\nmid y$.

Step 0: If $p \nmid a$, then $v_p(x^a-y^a) = v_p(x-y).$

To see this, write $\frac{x^a-y^a}{x-y} = x^{a-1} + x^{a-2}y + \cdots + y^{a-1}$, and since $x \equiv y$ mod $p$ this becomes

$x^{a-1} + x^{a-1} + \cdots + x^{a-1} \equiv ax^{a-1} \pmod p,$

which is nonzero since $p \nmid a$ and $p \nmid x$.

Step 1: Prove it for $n = p$.

In this case, $v_p(x^p-y^p) = v_p(x-y) + v_p\big(x^{p-1} + x^{p-2}y + \cdots + y^{p-1}\big)$, so the idea is to show that the latter term equals $v_p(p) = 1$.

To do this, write $y = x+pk$ for some $k$, and expand as a polynomial in $p$, looking mod $p^2$ $($throwing out terms with $p^2$ or higher$).$ Eventually we get

$\begin{aligned} x^{p-1} + x^{p-2}y + \cdots + y^{p-1} &\equiv x^{p-1} + \big(x^{p-1} + pkx^{p-2}\big) + \big(x^{p-1} + 2pkx^{p-2}\big) \\ &\phantom{\equiv} + \cdots + \big(x^{p-1}+(p-1)pkx^{p-2}\big) \\ &\equiv px^{p-1}+\frac{p(p-1)}2 pk x^{p-2} \\ &\equiv px^{p-1} \pmod {p^2}. \end{aligned}$

So it is divisible by $p$ but not $p^2$, as desired.

Step 2: Write $n = p^k a$, $p \nmid a$, and use the previous two steps repeatedly.

That is,

$\begin{aligned} v_p(x^n-y^n) &= v_p\left(\big(x^{p^k}\big)^a-\big(y^{p^k}\big)^a\right) \\ &= v_p\big(x^{p^k}-y^{p^k}\big) &&\text{(Step 0)} \\ &= v_p\left(\big(x^{p^{k-1}}\big)^p - \big(y^{p^{k-1}}\big)^p\right) \\ &= v_p\big(x^{p^{k-1}} - y^{p^{k-1}}\big) + 1 &&\text{(Step 1)} \end{aligned}$

and iterating the last two lines (or using induction) eventually gives $v_p(x-y) + k$, which is $v_p(x-y) + v_p(n)$. $_\square$