Mersenne Prime Exponent Cannot Be Even

Problem Statement

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

Hints

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

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

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

Solutions

Proof by Contrapositive:

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

Assume $n$ is even. Then $n = 2k$ for some integer $k$.

Since $n \geq 3$ and $n$ is even, we have $n \geq 4$, so $k \geq 2$.

Substitute:

Factor as difference of squares:

Since $k \geq 2$:

• $2^k \geq 4$, so $2^k - 1 \geq 3 > 1$
• $2^k \geq 4$, so $2^k + 1 \geq 5 > 1$

Both factors are integers strictly greater than $1$, so their product is composite.

Therefore, $2^n - 1$ is composite.

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

Note: This is why Mersenne primes have the form $2^p - 1$ where $p$ 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: https://vumaths.com/booklets/hsc-proofs/
• HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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