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Example 2

Find v5(1120−1)v_5(11^{20}-1).

We apply LTE with

p=5,a=11,b=1,n=20.p=5,\qquad a=11,\qquad b=1,\qquad n=20.

Since

5∣11−15\mid 11-1

and

5∤11,5\nmid 11,

LTE gives

v5(1120−1)=v5(11−1)+v5(20).v_5(11^{20}-1)=v_5(11-1)+v_5(20).

Now

v5(11−1)=v5(10)=1v_5(11-1)=v_5(10)=1

and

v5(20)=1.v_5(20)=1.

Therefore,

v5(1120−1)=2.v_5(11^{20}-1)=2.

So 2525 divides 1120−111^{20}-1, but 125125 does not.

Example 3

Find v7(1514−1)v_7(15^{14}-1).

Use LTE with

p=7,a=15,b=1,n=14.p=7,\qquad a=15,\qquad b=1,\qquad n=14.

Since

15≡1(mod7),15\equiv 1 \pmod 7,

we have

7∣15−1.7\mid 15-1.

Also 7∤157\nmid 15, so the conditions hold.

As such

v7(1514−1)=v7(15−1)+v7(14).v_7(15^{14}-1)=v_7(15-1)+v_7(14).

Now both 15−115-1 and 1414 contain exactly one factor of 77, so

v7(15−1)=1v_7(15-1)=1

and

v7(14)=1.v_7(14)=1.

Therefore,

v7(1514−1)=2.v_7(15^{14}-1)=2.

So the largest power of 77 dividing 1514−115^{14}-1 is 727^2.

Sum Form

There is also a useful sum version of LTE.

Let pp be an odd prime. Suppose that

p∣a+bp\mid a+b

and

p∤ab.p\nmid ab.

If nn is odd, then

vp(an+bn)=vp(a+b)+vp(n).v_p(a^n+b^n) = v_p(a+b)+v_p(n).

This follows by applying the difference form to

an−(−b)n.a^n-(-b)^n.

When nn is odd,

(−b)n=−bn,(-b)^n=-b^n,

so

an−(−b)n=an+bn.a^n-(-b)^n=a^n+b^n.

Worked Examples for the Sum Form

Example 4

Find v3(89+1)v_3(8^9+1).

We use the sum form with

p=3,a=8,b=1,n=9.p=3,\qquad a=8,\qquad b=1,\qquad n=9.

Since

3∣8+13\mid 8+1

and nn is odd, LTE gives

v3(89+1)=v3(8+1)+v3(9).v_3(8^9+1)=v_3(8+1)+v_3(9).

Now both 8+18+1 and 99 contain exactly two factors of 33, so

v3(8+1)=2v_3(8+1)=2

and

v3(9)=2.v_3(9)=2.

Therefore,

v3(89+1)=4.v_3(8^9+1)=4.

So the largest power of 33 dividing 89+18^9+1 is 343^4.

Example 5

Find v5(415+1)v_5(4^{15}+1).

Since

5∣4+15\mid 4+1

and 1515 is odd, we may apply the sum form:

v5(415+1)=v5(4+1)+v5(15).v_5(4^{15}+1)=v_5(4+1)+v_5(15).

Now

v5(4+1)=v5(5)=1v_5(4+1)=v_5(5)=1

and

v5(15)=1.v_5(15)=1.

Thus

v5(415+1)=2.v_5(4^{15}+1)=2.

So 2525 divides 415+14^{15}+1, but 125125 does not.

The Case p=2p=2

The prime 22 requires special care. The simplest useful version is the following.

If aa and bb are odd positive integers and nn is a positive integer, then

If nn is odd

v2(an−bn)=v2(a−b).v_2(a^n-b^n)=v_2(a-b).

If nn is even

v2(an−bn)=v2(a−b)+v2(a+b)+v2(n)−1.v_2(a^n-b^n) = v_2(a-b)+v_2(a+b)+v_2(n)-1.

This version is very useful, but it is also easier to misuse. Always check that aa and bb are odd before applying it.

Worked Examples for p=2p=2

Example 6

Find v2(512−1)v_2(5^{12}-1).

Here

a=5,b=1,n=12.a=5,\qquad b=1,\qquad n=12.

Since aa and bb are odd and nn is even, we use the even version:

v2(512−1)=v2(5−1)+v2(5+1)+v2(12)−1.v_2(5^{12}-1) = v_2(5-1)+v_2(5+1)+v_2(12)-1.

Now

v2(5−1)=v2(4)=2,v_2(5-1)=v_2(4)=2,
v2(5+1)=v2(6)=1,v_2(5+1)=v_2(6)=1,

and

v2(12)=2.v_2(12)=2.

Therefore,

v2(512−1)=2+1+2−1=4.v_2(5^{12}-1)=2+1+2-1=4.

So the largest power of 22 dividing 512−15^{12}-1 is 242^4.

Example 7

Find v2(79−1)v_2(7^9-1).

Here

a=7,b=1,n=9.a=7,\qquad b=1,\qquad n=9.

Since aa and bb are odd and nn is odd,

v2(79−1)=v2(7−1).v_2(7^9-1)=v_2(7-1).

Thus

v2(79−1)=v2(6)=1.v_2(7^9-1)=v_2(6)=1.

So the largest power of 22 dividing 79−17^9-1 is 22.

Using LTE in Olympiad Problems

LTE is most useful when a problem asks for:

  • The highest power of a prime dividing an expression
  • Divisibility by a large prime power
  • Solving equations involving expressions like an−bna^n-b^n
  • Comparing valuations on both sides of an equation
  • Proving that an expression is not a perfect power

A common pattern is to apply vpv_p to both sides of an equation.

For example, if a problem gives

an−bn=pk,a^n-b^n=p^k,

then applying LTE can turn the equation into a much simpler statement about

vp(a−b)+vp(n).v_p(a-b)+v_p(n).

Key Contest Insight

When you see an expression like

an−bna^n-b^n

or

an+bn,a^n+b^n,

and a prime divides a−ba-b or a+ba+b, consider using LTE.

The key question is not just whether the prime divides the expression, but how many times it divides the expression.

LTE converts a difficult divisibility question into a simple valuation formula.

Strategy Tips

  • Check whether the expression is of the form an−bna^n-b^n or an+bna^n+b^n.
  • Identify the prime pp whose power you want to count.
  • Check whether p∣a−bp\mid a-b for the difference form.
  • Check whether p∣a+bp\mid a+b and nn is odd for the sum form.
  • Make sure p∤abp\nmid ab.
  • Treat p=2p=2 separately.
  • After applying LTE, compute each valuation carefully.
  • In equation problems, take vpv_p of both sides.

Common Pitfalls

  • Applying the odd prime version when p=2p=2.
  • Forgetting to check that p∤abp\nmid ab.
  • Using the sum form when nn is even.
  • Confusing vp(a−b)v_p(a-b) with a−ba-b itself.
  • Forgetting that vp(n)v_p(n) contributes extra powers of pp.
  • Applying LTE before checking that pp divides a−ba-b or a+ba+b.

Practice Problems

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