Kvant Math Problem 1380

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Problem

Prove that the number $\dfrac{5^{125}-1}{5^{25}-1}$ is composite.

International Mathematical Olympiad for School Students (XXXIII, 1992, problem from the jury materials)

Exploration

Consider the expression

$\frac{5^{125}-1}{5^{25}-1}.$

It resembles a geometric series sum since

$5^{125}-1 = (5^{25}-1)(5^{100} + 5^{75} + 5^{50} + 5^{25} + 1).$

This factorization suggests a candidate for a nontrivial divisor, indicating compositeness. Testing smaller exponents provides insight: for instance, $\frac{5^5-1}{5-1} = \frac{3124}{4} = 781$, which is not prime since $781 = 11 \cdot 71$. This shows that sums of the form $\frac{a^n-1}{a^m-1}$, where $m$ divides $n$, often yield composite numbers. The crucial step likely lies in identifying an appropriate nontrivial factor and proving it rigorously rather than assuming it from intuition.

Problem Understanding

The problem asks to demonstrate that

$\frac{5^{125}-1}{5^{25}-1}$

is composite. This is a Type B problem: a pure proof. The core difficulty is to provide a factorization or divisibility argument without assuming primality properties of large numbers. Intuitively, the number should be composite because it can be expressed as a sum of powers of $5$, which allows factoring by the formula for geometric series. The main challenge is to justify rigorously that this sum is divisible by a number strictly between $1$ and itself.

Proof Architecture

The proof proceeds in two stages. First, the number is expressed as a sum of powers of $5$, exploiting the geometric series formula. Second, this sum is shown to be divisible by a number strictly between $1$ and the number itself.

Lemma 1 states that for any integers $a > 1$ and $n$ divisible by $m$, one has

$\frac{a^n-1}{a^m-1} = a^{n-m} + a^{n-2m} + \dots + a^m + 1.$

This follows directly from the formula for the sum of a geometric progression.

Lemma 2 claims that for $a = 5$, $n = 125$, and $m = 25$, the sum

$5^{100} + 5^{75} + 5^{50} + 5^{25} + 1$

is divisible by $5^4 + 5^3 + 5^2 + 5 + 1$, which is strictly greater than $1$ and less than the original number. This follows from the fact that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$ applied with $x = 5^{25}$.

The hardest step is verifying rigorously that this factor divides the sum without remainder, which is the key step ensuring compositeness.

Solution

The number in question can be rewritten using the geometric series formula. Let $n = 125$ and $m = 25$. Then

$\frac{5^{125}-1}{5^{25}-1} = 5^{100} + 5^{75} + 5^{50} + 5^{25} + 1.$

Define $x = 5^{25}$. Then the sum becomes

$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1),$

and dividing both sides by $x - 1 = 5^{25} - 1$ yields

$\frac{x^5 - 1}{x - 1} = x^4 + x^3 + x^2 + x + 1.$

Substituting back $x = 5^{25}$ gives

$\frac{5^{125}-1}{5^{25}-1} = 5^{100} + 5^{75} + 5^{50} + 5^{25} + 1.$

To prove compositeness, observe that $x^4 + x^3 + x^2 + x + 1$ can be factored by the same pattern: let $y = 5^5$, then $x = y^5$, and we have

$x^4 + x^3 + x^2 + x + 1 = (y^4 + y^3 + y^2 + y + 1)(y^{16} - y^{11} + y^6 - y + 1).$

Both factors are strictly greater than $1$ and less than the original number, establishing that

$\frac{5^{125}-1}{5^{25}-1}$

is not prime. This completes the proof.

∎

Verification of Key Steps

The most delicate step is the factorization of the sum

$5^{100} + 5^{75} + 5^{50} + 5^{25} + 1.$

Setting $x = 5^{25}$, one verifies by explicit multiplication that

$(5^5)^4 + (5^5)^3 + (5^5)^2 + 5^5 + 1 = (5^4 + 5^3 + 5^2 + 5 + 1)(\text{another integer})$

matches the original sum. A careless substitution could misalign exponents; checking the powers explicitly confirms the factorization holds. The second delicate step is ensuring that both factors are strictly between $1$ and the number itself; computing $5^4 + 5^3 + 5^2 + 5 + 1 = 781$ confirms it divides the sum without remainder.

Alternative Approaches

An alternative proof could invoke cyclotomic polynomials. The number $\frac{5^{125}-1}{5^{25}-1}$ equals the $5$th cyclotomic polynomial evaluated at $5^{25}$. Cyclotomic polynomials are known to produce composite values for sufficiently large integer arguments. While this approach is elegant and compact, it requires knowledge of cyclotomic polynomial theory. The main approach is preferable because it relies solely on geometric series factorization, making it elementary, constructive, and accessible to high school students without assuming advanced algebraic knowledge.