Lifting The Exponent Lemma
(LTE)

Amir Hossein Parvardi

Version 6
February 20, 2021

Lifting The Exponent Lemma is a powerful method for solving exponential Diophantine equations. It is pretty well-known in the Olympiad folklore (see, e.g., Reference 3) though its origins are hard to trace. Mathematically, it is a close relative of the classical Hensel’s lemma (see Reference 2) in number theory (in both the statement and the idea of the proof). In this article we analyze this method and present some of its applications.

We can use the Lifting The Exponent Lemma (this is a long name, let’s call it LTE!) in lots of problems involving exponential equations, especially when we have some prime numbers (and actually in some cases it “explodes” the problems). This lemma shows how to find the greatest power of a prime \(p\) —which is often \(\ge 3\)— that divides \(a^n+b^n\) for some positive integers \(a\) and \(b\). The proofs of theorems and lemmas in this article have nothing difficult and all of them use elementary mathematics. Understanding the theorem’s usage and its meaning is more important to you than remembering its detailed proof.

I have to thank Fedja, darij grinberg (Darij Grinberg), makar and ZetaX (Daniel) for their notifications about the article. And I specially appreciate JBL (Joel) and Fedja helps about \(\TeX\) issues.

Definitions and Notation

For two integers \(a\) and \(b\) we say \(a\) is divisible by \(b\) and write \(b|a\) if and only if there exists some integer \(q\) such that \(a=qb\).

We define \(v_p(x)\) to be the greatest power in which a prime \(p\) divides \(x\); in particular, if \(v_p(x)=\alpha\) then \(p^\alpha|x\) but \(p^{\alpha+1}\nmid x\). We also write \(p^\alpha \|x\), if and only if \(v_p(x)=\alpha\). So we have \(v_p(xy)=v_p(x)+v_p(y)\) and \(v_p(x+y)\ge\) \(\min\{v_p(x), v_p(y)\}\).

EXAMPLE

The greatest power of \(3\) that divides \(63\) is \(3^2\), because \(3^2=9|63\) but \(3^3=27\nmid 63\). In particular, \(3^2\|63\) or \(v_3(63)=2\).

EXAMPLE

Clearly we see that if \(p\) and \(q\) are two different prime numbers, then \(v_p(p^\alpha q^\beta)=\alpha\), or \(p^\alpha\|p^\alpha q^\beta\).

Note: We have \(v_p(0)=\infty\) for all primes \(p\).

Two Important and Useful Lemmas

Lemma 1

Let \(x\) and \(y\) be (not necessary positive) integers and let \(n\) be a positive integer. Given an arbitrary prime \(p\) (in particular, we can have \(p=2\)) such that \(\gcd(n,p)=1\), \(p|x−y\) and neither \(x\), nor \(y\) is divisible by \(p\) (i.e., \(p\nmid x\) and \(p\nmid y\)). We have

\[v_p(x^n−y^n)=v_p(x−y).\]

PROOF

We use the fact that

\[x^n−y^n=(x−y)\left(x^{n−1}+x^{n−2}y+x^{n−3} y^2+\cdots+y^{n−1}\right).\]

Now if we show that \(p\nmid x^{n−1}+x^{n−2}y\,+\) \(x^{n−3}y^2+\cdots+y^{n−1}\), then we are done.

In order to show this, we use the assumption \(p|x−y\). So we have \(x−y\equiv 0\;\bmod\;p\), or \(x\equiv y\;\bmod\;p\). Thus

\[\begin{align*} x^{n−1}&+x^{n−2}y+x^{n−3}y^2+\cdots+y^{n−1}\\ &\equiv x^{n−1}+x^{n−2}\cdot x+x^{n−3}\cdot x^2+\cdots+x\cdot x^{n−2}+x^{n−1}\\ &\equiv nx^{n−1}\\ &\not\equiv 0\;\bmod\;p. \end{align*}\]

This completes the proof.

Lemma 2

Let \(x\) and \(y\) be (not necessary positive) integers and let \(n\) be an odd positive integer. Given an arbitrary prime \(p\) (in particular, we can have \(p=2\)) such that \(\gcd(n,p)=1\), \(p|x+y\) and neither \(x\), nor \(y\) is divisible by \(p\), we have

\[v_p(x^n+y^n)=v_p(x+y).\]

PROOF

Since \(x\) and \(y\) can be negative, using Lemma 1 we obtain

\[v_p(x^n−(−y)^n)=v_p(x−(−y))\implies v_p(x^n+y^n)=v_p(x+y).\]

Note that since \(n\) is an odd positive integer we can replace \((−y)^n\) with \(−y^n\).

Lifting The Exponent Lemma (LTE)

Theorem 1 First Form of LTE

Let \(x\) and \(y\) be (not necessary positive) integers, let \(n\) be a positive integer, and let \(p\) be an odd prime such that \(p|x−y\) and none of \(x\) and \(y\) is divisible by \(p\) (i.e., \(p\nmid x\) and \(p\nmid y\)). We have

\[v_p(x^n−y^n)=v_p(x−y)+v_p(n).\]

PROOF

We may use induction on \(v_p(n)\). First, let us prove the following statement:

\[v_p(x^p−y^{\,p})=v_p(x−y)+1. \tag{1}\label{eq:1}\]

In order to prove this, we will show that

\[p | x^{p−1}+x^{p−2}y+\cdots+xy^{\,p−2}+y^{\,p−1}\;\tag{2}\label{eq:2}\]

and

\[p^2\nmid x^{p−1}+x^{p−2}y+\cdots+xy^{p−2}+y^{\,p−1}.\;\tag{3}\label{eq:3}\]

For \(\style{color:#5c5962;}{\eqref{eq:2}}\), we note that

\[x^{p−1}+x^{p−2}y+\cdots+xy^{\,p−2}+y^{\,p−1}\equiv px^{p−1}\equiv 0\mod p.\]

Now, let \(y=x+kp\), where \(k\) is an integer. For an integer \(1\le t< p\) we have

\[\begin{align*} y^{t}x^{p−1−t}&\equiv(x+kp)^{t}x^{p−1−t}\\ &\equiv x^{p−1−t}\left(x^t+t(kp)\left(x^{t−1}\right)+\frac{t(t-1)}{2}(kp)^2\left(x^{t−2}\right)+\cdots\right)\\ &\equiv x^{p−1−t}\left(x^t+t(kp)\left(x^{t-1}\right)\right) \\ &\equiv x^{p-1}+tkpx^{p-2}\mod p^2 \end{align*}\]

This means

\[y^tx^{p−1−t}\equiv x^{p−1}+tkpx^{p−2}\mod p^2,\;\;t=1, 2,3,4,\ldots,p−1.\]

Using this fact, we have

\[\begin{align*} x^{p−1}&+x^{p−2}y+\cdots+xy^{p−2}+y^{p−1}\\ &\equiv x^{p−1}+(x^{p−1}+kpx^{p−2})+(x^{p−1}+2kpx^{p−2})+\cdots+(x^{p−1}+(p-1)kpx^{p−2})\\ &\equiv px^{p−1}+(1+2+\cdots+p-1)kpx^{p−2}\\[0.5em] &\equiv px^{p−1}+\frac{p(p-1)}{2}kpx^{p−2}\\[0.5em] &\equiv px^{p−1}+\frac{p-1}{2}kp^2x^{p−1}\\[0.5em] &\equiv px^{p−1}\\ &\not\equiv 0\mod p^2.\\ \end{align*}\]

So we proved \(\style{color:#5c5962;}{\eqref{eq:3}}\) and the proof of \(\style{color:#5c5962;}{\eqref{eq:1}}\) is complete. Now let us return to our problem. We want to show that

\[v_p(x^n−y^n)=v_p(x−y)+v_p(n).\]

Suppose that \(n=p^\alpha b\) where \(\gcd(p,b)=1\). Then

\[\begin{align*} v_p(x^n−y^n)&=v_p\left(\left(x^{p^\alpha}\right)^b−\left(y^{\,p^\alpha}\right)^b\right)\\ &=v_p\left(x^{p^\alpha}−y^{\,p^\alpha}\right)=v_p\left(\left(x^{p^{\alpha-1}}\right)^p−\left(y^{\,p^{\alpha-1}}\right)^p\right)\\ &=v_p\left(x^{p^{\alpha-1}}−y^{\,p^{\alpha-1}}\right)+1=v_p\left(\left(x^{p^{\alpha-2}}\right)^p−\left(y^{\,p^{\alpha-2}}\right)^p\right)+1\\ &=v_p\left(x^{p^{\alpha-2}}−y^{\,p^{\alpha−2}}\right)+2\\ &\:\;\vdots\\ &=v_p\left(x^{p^1}−y^{\,p^1}\right)+\alpha−1=v_p(x-y)+\alpha\\ &=v_p(x-y)+v_p(n). \end{align*}\]

Note that we used the fact that if \(p|x−y\), then we have \(p|x^k−y^k\), because we have \(x−y|x^k−y^k\) for all positive integers \(k\). The proof is complete.

Theorem 2 Second Form of LTE

Let \(x\), \(y\) be two integers, \(n\) be an odd positive integer, and \(p\) be an odd prime such that \(p|x+y\) and none of \(x\) and \(y\) is divisible by \(p\). We have

\[v_p(x^n+y^n)=v_p(x+y)+v_p(n).\]

PROOF

This is obvious using Theorem 1. See the trick we used in proof of Lemma 2.

What about \(p=2\)?

Question. Why did we assume that \(p\) is an odd prime, i.e., \(p\ne 2\)? Why can’t we assume that \(p=2\) in our proofs?

Hint. Note that \(\frac{p−1}{2}\) is an integer only for \(p>2\).

Theorem 3 LTE for the case \(p=2\)

Let \(x\) and \(y\) be two odd integers such that \(4| x−y\). Then

\[v_2\left(x^n−y^n\right)=v_2(x−y)+v_2(n).\]

PROOF

We showed that for any prime \(p\) such that \(\gcd(p,n)= 1\), \(p|x−y\) and none of \(x\) and \(y\) is divisible by \(p\), we have

\[v_p(x^n-y^n)=v_p(x−y)\]

Let \(m\) be an odd integer and \(k\) be a positive integer such that \(n=m\cdot2^k\). So it suffices to show that

\[v_2\left(x^{2^k}−y^{2^k}\right) =v_2(x−y)+k.\]

Factorization gives

\[x^{2^k}−y^{2^k}=\left(x^{2^{k−1}}+ y^{2^{k−1}}\right)\left(x^{2^{k−2}}+ y^{2^{k−2}}\right)\cdots(x^2+y^2)(x+y)(x-y)\]

Now since \(x\equiv y\equiv\pm 1\;\bmod\;4\) then we have \(x^{2^i}\equiv\) \(y^{2^i}\equiv 1\;\bmod\;4\) for all positive integers \(i\) and so \(x^{2^i}+y^{2^i}\equiv 2\;\bmod\;4,\) \(i=1,2,3,\ldots\) Also, since \(x\) and \(y\) are odd and \(4|x−y\), we have \(x+y\equiv 2\;\bmod\;4\). This means the power of \(2\) in all of the factors in the above product (except \(x−y\)) is one. We are done.

Theorem 4

Let \(x\) and \(y\) be two odd integers and let \(n\) be an even positive integer. Then

\[v_2(x^n−y^n)=v_2(x−y)+v_2(x+y)+v_2(n)−1.\]

PROOF

We know that the square of an odd integer is of the form \(4k+1\). So for odd \(x\) and \(y\) we have \(4|x^2−y^2\). Now let \(m\) be an odd integer and \(k\) be a positive integer such that \(n=m\cdot2^k\). Then

\[\begin{align*} v_2(x^n−y^n)&=v_2\left(x^{m\cdot2^k}−y^{m\cdot2^k}\right)\\ &=v_2\left(\left(x^2\right)^{2^{k−1}}−\left(y^2\right)^{2^{k−1}}\right)\\ &\;\;\vdots\\ &=v_2\left(x^2−y^2\right)+k−1\\ &=v_2(x−y)+v_2(x+y)+v_2(n)−1. \end{align*}\]

Summary

Let \(p\) be a prime number and let \(x\) and \(y\) be two (not necessary positive) integers that are not divisible by \(p\). Then:

a. For a positive integer \(n\)

• if \(p\ne 2\) and \(p|x−y\), then

\[v_p(x^n−y^n)=v_p(x−y)+v_p(n).\]

• if \(p=2\) and \(4|x−y\), then

\[v_2(x^n−y^n)=v_2(x−y)+v_2(n).\]

• if \(p=2\), \(n\) is even, and \(2|x−y\), then

\[v_2(x^n−y^n)=v_2(x-y)+v_2(x+y)+v_2(n)−1.\]

b. For an odd positive integer \(n\), if \(p|x+y\), then

\[v_p(x^n+y^n)=v_p(x+y)+v_p(n).\]

c. For a positive integer \(n\) with \(\gcd(p,n)=1\), if \(p|x−y\), we have

\[v_p(x^n-y^n)=v_p(x-y).\]

If \(n\) is odd, \(\gcd(p,n)=1\), and \(p|x+y\), then we have

\[v_p(x^n+y^n)=v_p(x+y).\]

Note: The most common mistake in using LTE is when you don’t check the \(p|x\pm y\) condition, so always remember to check it. Otherwise your solution will be completely wrong.

Problems with solutions

Problem 1 Russia 1996

Find all positive integers \(n\) for which there exist positive integers \(x\), \(y\) and \(k\) such that \(\gcd(x,y)=1\), \(k>1\) and \(3^n=x^k+y^k\).

SOLUTION

\(k\) should be an odd integer (otherwise, if \(k\) is even, then \(x^k\) and \(y^k\) are perfect squares, and it is well known that for integers \(a\), \(b\) we have \(3|a^2+b^2\) if and only if \(3|a\) and \(3|b\), which is in contradiction with \(\gcd(x,y)=1\)).

Suppose that there exists a prime \(p\) such that \(p|x+y\). This prime should be odd. So \(v_p(3^n)=v_p(x^k+y^k)\), and using Theorem 2 we have \(v_p(3^n)=\) \(v_p(x^k+y^k)=\) \(v_p(k)+v_p(x+y)\). But \(p|x+y\) means that \(v_p(x+y)\ge 1>0\) and so \(v_p(3^n)=v_p(k)+v_p(x+y)>0\) and so \(p|3^n\). Thus \(p=3\). This means \(x+y=3^m\) for some positive integer \(m\). Note that \(n=v_3(k)+m\). There are two cases:

• \(m>1\)

  We can prove by induction that \(3^a\ge a+2\) for all integers \(a\ge 1\), and so we have \(v_3(k)\le k − 2\) (why?). Let \(M=\max(x,y)\). Since \(x+y=3^m\ge 9\), we have \(M\ge 5\). Then

  \[\begin{align*} M^{k-1}\ge 5^{k-1}&\land M\ge \frac{x+y}{2}=\frac{1}{2}\cdot 3^m\\[1em] x^k+y^k&\ge M^k=M\cdot M^{k-1}\\ &\phantom{\ge}>\frac{1}{2}\cdot 3^m\cdot5^{k-1}\\ &\phantom{\ge >}>3^m\cdot 5^{k-2}\\ &\phantom{\ge >>}\ge 3^{m+k-2}\\ &\phantom{\ge >>\ge}\ge 3^{m+v_3(k)}=3^n \end{align*}\]

  which is a contradiction.

• \(m=1\)

  Then \(x+y=3\), so \(x=1\), \(y=2\) (or \(x=2\), \(y=1\)). Thus \(3^{1+v_3(k)}=1+2^k\). But note that \(3^{v_3(k)}|k\) so \(3^{v_3(k)}\le k\). Thus

  \[1+2^k=3^{v_3(k)+1}=3\cdot3^{v_3(k)}\le 3k\implies 1+2^k\le 3k\]

  And one can check that the only odd value of \(k>1\) that satisfies the above inequality is \(k=3\). So \((x,y,n,k)=\) \((1, 2, 2, 3)\), \((2, 1, 2, 3)\) in this case.

Thus, the final answer is \(n=2\).

Problem 2 Balkan 1993

Let \(p\) be a prime number and \(m>1\) be a positive integer. Show that if for some positive integers \(x>1\), \(y>1\) we have

\[\frac{x^p+y^{\,p}}{2}=\left(\frac{x+y}{2}\right)^m,\]

then \(m=p\).

SOLUTION

One can prove by induction on \(p\) that \(\frac{x^p+y^{\!\;p}}{2}\ge\left(\frac{x+y}{2}\right)^p\) for all positive integers \(p\). Now since \(\frac{x^p+y^{\!\;p}}{2}=\left(\frac{x+y}{2}\right)^m\), we should have \(m\ge p\). Let \(d=\gcd(x,y)\), so there exist positive integers \(x_1\), \(y_1\) with \(\gcd(x_1,y_1)=1\) such that \(x=dx_1\), \(y=dy_1\) and \(2^{m−1}(x^p_1+y^{\,p}_1)=d^{m−p}(x_1+y_1)^m\). There are two cases:

• Assume that \(p\) is odd.

  Take any prime divisor \(q\) of \(x_1+y_1\) and let \(v=v_q(x_1+y_1)\). If \(q\) is odd, we see that \(v_q(x^{p}_1+y^{\,p}_1)=v+v_q(p)\) and \(v_q(d^{m−p}(x_1+y_1)^m)\ge mv\) (because \(q\) may also be a factor of \(d\)). Thus \(m\le 2\) and \(p\le 2\), giving an immediate contradiction. If \(q=2\), then \(m-1+v\ge mv\), so \(v\le 1\) and \(x_1+y_1=2\), i.e., \(x=y\), which immediately implies \(m=p\).

• Assume that \(p=2\).

  We notice that for \(x+y\ge 4\) we have \(\frac{x^2+y^2}{2}<\) \(2\left(\frac{x+y}{2}\right)^2\le\) \(\left(\frac{x+y}{2}\right)^3\), so \(m=2\). It remains to check that the remaining cases \((x,y)=(1,2),(2,1)\) are impossible.

Problem 3

Find all positive integers \(a\), \(b\) that are greater than \(1\) and satisfy

\[b^a|a^b−1.\]

SOLUTION

Let \(p\) be the least prime divisor of \(b\). Let \(m\) be the least positive integer for which \(p|a^m−1\). Then \(m|b\) and \(m|p−1\), so any prime divisor of \(m\) divides \(b\) and is less than \(p\). Thus, not to run into a contradiction, we must have \(m=1\). Now, if \(p\) is odd, we have \(av_p(b)\le v_p(a − 1)+v_p(b)\), so \((a−1)\) \(\le (a−1)v_p(b)\) \(\le v_p(a−1)\), which is impossible. Thus \(p=2\), \(b\) is even, \(a\) is odd and \(av_2(b)\le v_2(a−1)\,+\,\)​\(v_2(a+1)+v_2(b)−1\) whence \(a\le\) \((a−1)v_2(b)+1\le\) \(v_2(a − 1)\,+\,\)​\(v_2(a+1)\), which is possible only if \(a=3\), \(v_2(b)=1\). Put \(b=2B\) with odd \(B\) and rewrite the condition as \(2^3B^3|3^{2B}−1\). Let \(q\) be the least prime divisor of \(B\) (now, surely, odd). Let \(n\) be the least positive integer such that \(q|3^n−1\). Then \(n|2B\) and \(n|q−1\) whence \(n\) must be \(1\) or \(2\) (or \(B\) has a smaller prime divisor), so \(q|3−1=2\) or \(q|3^2−1=8\), which is impossible.

Thus \(B=1\) and \(b=2\).

Problem 4

Find all positive integer solutions of the equation \(x^{2009}+y^{2009}=7^z\).

SOLUTION

Factor \(2009\). We have \(2009=7^2\cdot 41\). Since \(x+y|x^{2009}+y^{2009}\) and \(x+y>1\), we must have \(7| x+y\). Removing the highest possible power of \(7\) from \(x\), \(y\) (why?), we get a solution of the equation such that \(7\nmid x,y\). We only need to check if there exists that solution. In that case, \(v_7(x^{2009}+y^{2009})=\) \(v_7(x+y)+\) \(v_7(2009)=\) \(v_7(x+y)+2\), so \(x^{2009}+y^{2009}=\) \(49\cdot k\cdot(x+y)\) where \(7\nmid k\). But we have \(x^{2009}+y^{2009}=7^z\), which means the only prime factor of \(x^{2009}+y^{2009}\) is \(7\), so \(k=1\). Thus \(x^{2009}+y^{2009}=\) \(49(x+y)\). But in this equation the left hand side is much larger than the right hand one if \(\max(x,y)>1\), and, clearly, \((x,y)=(1,1)\) is not a solution. Thus the given equation does not have any solutions in the set of positive integers.

Challenge Problems

Problem 1

Let \(k\) be a positive integer. Find all positive integers \(n\) such that \(3^k|2^n−1\).

Problem 2 UNESCO Competition 1995

Let \(a\), \(n\) be two positive integers and let \(p\) be an odd prime number such that

\[a^p\equiv 1\mod{p^n}\]

Prove that

\[a\equiv 1\mod{p^{n-1}}\]

Problem 3 Iran Second Round 2008

Show that the only positive integer value of \(a\) for which \(4(a^n+1)\) is a perfect cube for all positive integers \(n\), is \(1\).

Problem 4

Let \(k>1\) be an integer. Show that there exists infinitely many positive integers \(n\) such that

\[n|1^n+2^n+3^n+\cdots+k^n.\]

Problem 5 Ireland 1996

Let \(p\) be a prime number, and \(a\) and \(n\) positive integers. Prove that if

\[2^p+3^p=a^n\]

then \(n=1\).

Problem 6 Russia 1996

Let \(x\), \(y\), \(p\), \(n\), \(k\) be positive integers such that \(n\) is odd and \(p\) is an odd prime. Prove that if \(x^n+y^n=p^k\), then \(n\) is a power of \(p\).

Problem 7

Find the sum of all the divisors \(d\) of \(N=19^{88}−1\) which are of the form \(d=2^a3^b\) with \(a,b\in\mathbb{N}\).

Problem 8

Let \(p\) be a prime number. Solve the equation \(a^{p}−1 =p^k\) in the set of positive integers.

Problem 9

Find all solutions of the equation

\[(n−1)!+1=n^m\]

in positive integers.

Problem 10 Bulgaria 1997

For some positive integer \(n\), the number \(3^n−2^n\) is a perfect power of a prime. Prove that \(n\) is a prime.

Problem 11

Let \(m\), \(n\), \(b\) be three positive integers with \(m\ne n\) and \(b>1\). Show that if prime divisors of the numbers \(b^n−1\) and \(b^m−1\) be the same, then \(b+1\) is a perfect power of \(2\).

Problem 12 IMO ShortList 1991

Find the highest degree \(k\) of \(1991\) for which \(1991^k\) divides the number

\[1990^{1991^{1992}}+1992^{1991^{1990}}.\]

Problem 13

Prove that the number \(a^{a−1}−1\) is never square-free for all integers \(a>2\).

Problem 14 Czech Slovakia 1996

Find all positive integers \(x\), \(y\) such that \(p^x−y^{\,p}=1\), where \(p\) is a prime.

Problem 15

Let \(x\) and \(y\) be two positive rational numbers such that for infinitely many positive integers \(n\), the number \(x^n−y^n\) is a positive integer. Show that \(x\) and \(y\) are both positive integers.

Problem 16 IMO 2000

Does there exist a positive integer \(n\) such that \(n\) has exactly \(2000\) prime divisors and \(n\) divides \(2^n+1\)?

Problem 17 China Western Mathematical Olympiad 2010

Suppose that \(m\) and \(k\) are non-negative integers, and \(p=2^{2^{m}}+1\) is a prime number. Prove that

• \(2^{2^{m+1}p^k}\equiv 1\mod p^{k+1}\)
• \(2^{m+1}p^k\) is the smallest positive integer \(n\) satisfying the congruence equation \(2^n\equiv 1\mod p^{k+1}\).

Problem 18

Let \(p\ge 5\) be a prime. Find the maximum value of positive integer \(k\) such that

\[p^k|(p−2)^{2(p−1)}−(p−4)^{p−1}.\]

Problem 19

Let \(a\), \(b\) be distinct real numbers such that the numbers

\[a−b,a^2−b^2,a^3−b^3,\ldots\]

are all integers. Prove that \(a\), \(b\) are both integers.

Problem 20 MOSP 2001

Find all quadruples of positive integers \((x,r,p,n)\) such that \(p\) is a prime number, \(n,r>1\) and \(x^r−1=p^n\).

Problem 21 China TST 2009

Let \(a>b>1\) be positive integers and \(b\) be an odd number, let \(n\) be a positive integer. If \(b^n|a^n−1\), then show that \(a^b>\frac{3^n}{n}\).

Problem 22 Romanian Junior Balkan TST 2008

Let \(p\) be a prime number, \(p\ne 3\), and integers \(a\), \(b\) such that \(p|a+b\) and \(p^2|a^3+b^3\). Prove that \(p^2|a+b\) or \(p^3|a^3+b^3\).

Problem 23

Let \(m\) and \(n\) be positive integers. Prove that for each odd positive integer \(b\) there are infinitely many primes \(p\) such that \(p^n\equiv 1\!\!\mod b^m\) implies \(b^{m−1}|n\).

Problem 24 IMO 1990

Determine all integers \(n>1\) such that

\[\frac{2^{n} + 1}{n^2}\]

is an integer.

Problem 25

Find all positive integers \(n\) such that

\[\frac{2^{n-1} + 1}{n}\]

is an integer.

Problem 26

Find all primes \(p\), \(q\) such that \(\dfrac{(5^p−2^p)(5^q−2^q)}{pq}\) is an integer.

Problem 27

For some natural number \(n\) let \(a\) be the greatest natural number for which \(5^n−3^n\) is divisible by \(2^a\). Also let \(b\) be the greatest natural number such that \(2^b\le n\). Prove that \(a\le b+3\).

Problem 28

Determine all sets of non-negative integers \(x\), \(y\) and \(z\) which satisfy the equation

\[2^x+3^y=z^2.\]

Problem 29 IMO ShortList 2007

Find all surjective functions \(f : \mathbb{N}\rightarrow \mathbb{N}\) such that for every \(m,n\in\mathbb{N}\) and every prime \(p\), the number \(f(m+n)\) is divisible by \(p\) if and only if \(f(m)+f(n)\) is divisible by \(p\).

Problem 30 Romania TST 1994

Let \(n\) be an odd positive integer. Prove that \(\left((n−1)^n+1\right)^2\) divides \(n(n−1)^{(n−1)^n+1}+n\).

Problem 31

Find all positive integers \(n\) such that \(3^n−1\) is divisible by \(2^n\).

Problem 32 Romania TST 2009

Let \(a,n\ge 2\) be two integers, which have the following property: there exists an integer \(k\ge 2\), such that \(n\) divides \((a−1)^k\). Prove that \(n\) also divides

\[a^{n−1}+a^{n−2}+\cdots+a+1.\]

Problem 33

Find all the positive integers \(a\) such that \(\frac{5^a+1}{3^a}\) is a positive integer.

Hints and Answers to Selected Problems

Problem 1

Answer: \(n=2\cdot 3^{k−1}s\) for some \(s\in\mathbb{N}\).

Problem 2

Show that \(v_p(a−1)=v_p(a^p−1)−1\ge\)\(n−1\).

Problem 3

If \(a>1\), \(a^2+1\) is not a power of \(2\) (because it is \(> 2\) and either \(1\) or \(2\) modulo \(4\)). Choose some odd prime \(p|a^2+1\). Now, take some \(n=2m\) with odd \(m\) and notice that \(v_p(4(a^n+1))=\) \(v_p(a^2+1)+v_p(m)\) but \(v_p(m)\) can be anything we want modulo \(3\).

Problem 5

\(2^p+3^p\) is not a square, and use the fact that \(v_5(2^p+3^p)=\) \(1+v_5(p) \le 2\).

Problem 8

Consider two cases: \(p=2\) and \(p\) is an odd prime. The latter does not give any solutions.

Problem 9

\((n,m)=(2,1)\) is a solution. In other cases, show that \(n\) is an odd prime and \(m\) is even. The other solution is \((n,m)=(5,2)\).

Problem 12

Answer: \(\max(k)=1991\).

Problem 13

Take any odd prime \(p\) such that \(p|a−1\). It’s clear that \(p^2|a^{a−1}−1\).

Problem 14

Answer: \((p,x,y)=(2, 1, 1),\) \((3, 2, 1)\).

Problem 18

Let \(p−1=2^sm\) and show that \(v_p(2^{s−1}m)=0\). The maximum of \(k\) is \(1\).

Problem 19

Try to prove Problem 15 first.

Problem 20

Show that \(p=2\) and \(r\) is an even positive integer.

Problem 22

If \(p|a\), \(p|b\), then \(p^3|a^3+b^3\). Otherwise LTE applies and \(v_p(a+b)=\) \(v_p(a^3+b^3)\ge 2\).

Problem 24

The answer is \(n=1\) or \(n=3\).

Problem 26

Answer: \((p,q)=(3,3),(3,13)\).

Problem 27

If \(n\) is odd, then \(a=1\). If \(n\) is even, then \(a=v_2(5^n−3^n)=\) \(v_2(5−3)+v_2(5+3)+v_2(n)−1=\) \(3+v_2(n)\). But, clearly, \(b\ge v_2(n)\).

Problem 30

\(n|(n−1)^n+1\), so for every \(p|(n−1)^n+1\), we have

\[\begin{align*} v_p\left((n−1)^{(n−1)^n+1}+1\right)&= v_p((n − 1)^n+1)+v_p\left(\frac{(n−1)^{n+1}+1}{n}\right)\\ &=2v_p((n−1)^n+1)−v_p(n) \end{align*}\]

which completes the proof.

Problem 31

\(n\le v_2(3^n−1)\le 3+v_2(n)\), so \(n\le 4\).

Problem 33

\(a\) must be odd (otherwise the numerator is \(2\;\bmod\;3\)). Then \(a\le v_3(5^a+1)=\) \(1+v_3(a)\) giving \(a=1\) as the only solution.