This is copy/pasted from a submitted coauthored paper of mine:

Lemma: Let $k \ge 1$ and $p$ be prime. If $a \equiv b \pmod p$, $c \ge 0$ and $c \equiv 0 \pmod{p^{k-1}}$, then $a^c \equiv b^c \pmod {p^{k}}$.

A reviewer pointed out that this is called lifting the exponent (I had no idea). Indeed, I found that it's a consequence of the Second Form of LTE on the Art of Problem Solving:

Second Form of LTE: Let $ x,y$ be two integers, $n$ be an odd positive integer, and $p$ be an odd prime such that $ p\mid x + y$ and none of $x$ and $y$ are divisible by $p.$ then: $$v_p( x^n + y^n ) =v_p( x + y ) + v_p( n ).$$

(Here $v_p(m)$ is the largest non-negative integer for which $p^{v_p(m)}$ divides $m$.)

I'd like to add a reference to our paper, but I'm not sure where to look. Searching for "lifting the exponent" in Google Books gives a single conference proceedings book containing the result, but in some completely irrelevant paper. Similarly, searching for "lifting the exponent" in Google Scholar gives places its mentioned which are completely irrelevant our paper.

Question: What is a good quality reference for "lifting the exponent"?

I guess I'm looking for a reference in a number theory textbook; something better than a random paper which mentions it.


I don't know if the OP is still interested in the question after so long time, but the second form of LTE is typically attributed to É. Lucas:

É. Lucas, Théorie des Fonctions Numériques Simplement Periodiques, Amer. J. Math. 1 (1878), 184–196, 197-240, 289–321,

and R. D. Carmichael:

R. D. Carmichael, On the Numerical Factors of Certain Arithmetic Forms, Amer. Math. Monthly 16 (1909), no. 10, 153–159.

In particular, Carmichael fixed a mistake in Lucas' original work in the $2$-adic case.

| cite | improve this answer | |
  • $\begingroup$ Thank you. Our paper [open access] was already published, and we just wrote "Lemma 2 is a case of a folklore result known as “lifting the exponent.”" But I'll know where to look if it comes up again. $\endgroup$ – Rebecca J. Stones Jan 16 '17 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.