Mersenne Prime Exponent Cannot Be Even

#hsc-maths#proofs#medium

In this problem we explore a classic property of Mersenne primes: if 2n12^n-1 is prime, then nn must be odd—in fact, prime. This is a great exercise in proof by contrapositive and algebraic factorisation, techniques that appear frequently in the HSC Mathematics Extension 1 course. By working through the solution, you'll see how a simple difference-of-squares factorisation can transform a statement about primality into an argument about composite numbers.

Problem Statement

Prove that for all integers n3n \geq 3, if 2n12^n - 1 is prime, then nn cannot be even.


Hints

Use proof by contrapositive. Instead of proving "P    QP \implies Q", prove "¬Q    ¬P\neg Q \implies \neg P".

That is, prove: "If nn is even (and n3n \geq 3), then 2n12^n - 1 is composite."

Write n=2kn = 2k and factor 22k12^{2k} - 1 as a difference of squares.


Solutions

Proof by Contrapositive:

We prove the contrapositive: If nn is even (with n3n \geq 3), then 2n12^n - 1 is composite.

Assume nn is even. Then n=2kn = 2k for some integer kk.

Since n3n \geq 3 and nn is even, we have n4n \geq 4, so k2k \geq 2.

Now, 2n12^n - 1 becomes 22k12^{2k} - 1. Using the index law 22k=(2k)22^{2k} = (2^k)^2, we can rewrite the expression as a difference of two squares:

2n1=22k1=(2k)2122^n - 1 = 2^{2k} - 1 = (2^k)^2 - 1^2

Factor as difference of squares:

2n1=(2k1)(2k+1)2^n - 1 = (2^k - 1)(2^k + 1)

Since k2k \geq 2:

  • 2k42^k \geq 4, so 2k13>12^k - 1 \geq 3 > 1
  • 2k42^k \geq 4, so 2k+15>12^k + 1 \geq 5 > 1

Both factors are integers strictly greater than 11, so their product is composite.

Therefore, 2n12^n - 1 is composite.

By contrapositive, if 2n12^n - 1 is prime, then nn cannot be even. \blacksquare

Note: This is why Mersenne primes have the form 2p12^p - 1 where pp itself is prime (though not all such numbers are prime).


Takeaways

  • Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
  • Check edge cases and verify where each assumption is used in the argument.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Polynomials, HSC Trigonometry

Written by Vu Hung Nguyen

Mathematics Educator · LinkedIn · About