Mersenne Prime Exponent Cannot Be Even
In this problem we explore a classic property of Mersenne primes: if is prime, then 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 , if is prime, then cannot be even.
Hints
Use proof by contrapositive. Instead of proving "", prove "".
That is, prove: "If is even (and ), then is composite."
Write and factor as a difference of squares.
Solutions
Proof by Contrapositive:
We prove the contrapositive: If is even (with ), then is composite.
Assume is even. Then for some integer .
Since and is even, we have , so .
Now, becomes . Using the index law , we can rewrite the expression as a difference of two squares:
Factor as difference of squares:
Since :
- , so
- , so
Both factors are integers strictly greater than , so their product is composite.
Therefore, is composite.
By contrapositive, if is prime, then cannot be even.
Note: This is why Mersenne primes have the form where 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