Evenness Equivalence for x and x Squared

Problem Statement

Prove that $x$ is even if and only if $x^2$ is even (for integer $x$).

Hints

This is a biconditional (iff) statement requiring two proofs:

• Direction 1: If $x$ is even, then $x^2$ is even (direct proof).
• Direction 2: If $x^2$ is even, then $x$ is even (try contrapositive: if $x$ is odd, then $x^2$ is odd).

Solutions

Direction 1 ($\Rightarrow$): If $x$ is even, then $x^2$ is even.

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

Therefore:

Since $2k^2$ is an integer, $x^2$ is even by definition.

Direction 2 ($\Leftarrow$): If $x^2$ is even, then $x$ is even.

We prove the contrapositive: If $x$ is odd, then $x^2$ is odd.

Assume $x$ is odd. Then $x = 2k + 1$ for some integer $k$.

Therefore:

Since $2k^2 + 2k$ is an integer, $x^2$ is odd by definition.

By contrapositive, if $x^2$ is even, then $x$ must be even.

Conclusion: Both directions proven, so $x$ is even $\iff$ $x^2$ is even. $\blacksquare$

Note: This result is fundamental in many irrationality proofs (e.g., $\sqrt{2}$ is irrational). If $p$ is an odd prime and $p|x^2$, then $p|x$.

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 Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/
• HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/

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