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 squarefree integer. Show that there is no pair of coprime positive integers \( (x,y)\) such that \[(x+y)^3 | (x^n+y^n).\]
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, \[ v_p(n) = k \Leftrightarrow p^k|n \ \text{and} \ p^{k+1} \nmid n. \]

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, \[ v_3(5^{18}-2^{18}) = v_3(5-2)+v_3(18) = 1+2 = 3. \] So the answer is \( 3^3\).

Without LTE, the problem can be solved by factoring \[ \begin{align} 5^{18}-2^{18}&=\big(5^9-2^9\big)\big(5^9+2^9\big) \\ &= \big(5^3-2^3\big)\big(2^6+2^35^3+5^6\big)\big(5^9+2^9\big) \\ &= (5-2)\big(2^2+(2)(5)+5^2\big)\big(2^6+2^35^3+5^6\big)\big(5^9+2^9\big). \end{align} \] 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.

Assume \( (x+y)^3|(x^n+y^n) \) 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 \( 8 \), so \( v_2(x^n+y^n) = 1 \), 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. \(_\square \)

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

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 \[ \begin{align} x^{k-1} - x^{k-2}y + \cdots - xy^{k-2} + y^{k-1} &= k \\ (x-y)\big(x^{k-2} +x^{k-4}y^2+ \cdots + xy^{k-3}\big) + y^{k-1} &=k. \end{align} \] 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. \(_\square \)

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 \[ z^n | (x^n-y^n) \ \text{for all} \ n. \] 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 \( a = \dfrac{\hspace{1.5mm}\frac xp\hspace{1.5mm}}{\hspace{1.5mm}\frac zp\hspace{1.5mm}}, b = \dfrac{\hspace{1.5mm}\frac yp\hspace{1.5mm}}{\hspace{1.5mm}\frac zp\hspace{1.5mm}} \) 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 \[ \begin{align} p^n &\le (x-y)n \\ \frac{p^n}{n} &\le (x-y) \end{align} \] 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 \[ \begin{align} x^{p-1} + x^{p-2}y + \cdots + y^{p-1} &\equiv x^{p-1} + \big(x^{p-1} + pkx^{p-2}\big) + \big(x^{p-1} + 2pkx^{p-2}\big) \\ &\phantom{\equiv} + \cdots + \big(x^{p-1}+(p-1)pkx^{p-2}\big) \\ &\equiv px^{p-1}+\frac{p(p-1)}2 pk x^{p-2} \\ &\equiv px^{p-1} \pmod {p^2}. \end{align} \] So it is divisible by \( p \) but not \( p^2 \), as desired.

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

That is, \[ \begin{align} v_p(x^n-y^n) &= v_p\left(\big(x^{p^k}\big)^a-\big(y^{p^k}\big)^a\right) \\ &= v_p\big(x^{p^k}-y^{p^k}\big) &&\text{(Step 0)} \\ &= v_p\left(\big(x^{p^{k-1}}\big)^p - \big(y^{p^{k-1}}\big)^p\right) \\ &= v_p\big(x^{p^{k-1}} - y^{p^{k-1}}\big) + 1 &&\text{(Step 1)} \end{align} \] and iterating the last two lines (or using induction) eventually gives \( v_p(x-y) + k \), which is \( v_p(x-y) + v_p(n) \). \(_\square \)