Prove That If x + 5 Is Odd, Then x² Is Even

Assume that $x$ is an integer. The statement is then true: if $x+5$ is odd, $x$ must be even, and the square of an even integer is even. The proof follows directly from the definitions of odd and even.

Direct proof from the definitions

An integer is odd if it can be written as $2k+1$ for some integer $k$. If $x+5$ is odd, there is an integer $k$ such that

$$x+5=2k+1.$$

Subtract $5$ from both sides:

$$\begin{aligned} x &= 2k+1-5 \\ &= 2k-4 \\ &= 2(k-2). \end{aligned}$$

Because $k$ is an integer, $k-2$ is also an integer. Thus $x$ has the form $2m$ for the integer $m=k-2$, so $x$ is even. Squaring gives

$$x^2 = [2(k-2)]^2 = 4(k-2)^2 = 2\bigl(2(k-2)^2\bigr).$$

The quantity $2(k-2)^2$ is an integer, so $x^2$ is twice an integer. That is exactly the definition of an even integer. This proves the claim.

A shorter parity argument

Adding an odd integer to an integer changes its parity: an even integer plus an odd integer is odd, while an odd integer plus an odd integer is even. Since $5$ is odd, the only way for $x+5$ to be odd is for $x$ to be even. The square of an even integer is even, because if $x=2m$, then $x^2=4m^2=2(2m^2)$.

The definition-based proof is often preferable in a formal proof because it shows exactly which integer witnesses make the expressions even or odd.

Worked example

Take $x=8$. Then $x+5=13$, which is odd. The proof’s representation applies with $k=6$, since $13=2(6)+1$. It gives

$$x=2(6-2)=8,$$

so $x$ is even. Its square is $x^2=64=2\cdot 32$, also even.

The same reasoning works for negative integers. For instance, if $x=-4$, then $x+5=1$ is odd, and $x^2=16$ is even. The definitions use integer multiples of $2$ and do not require the integer to be positive.

Why the integer assumption matters

Odd and even are normally defined only for integers. If $x$ is an arbitrary real number, the phrases “$x$ is odd” and “$x$ is even” do not have their usual number-theoretic meaning. The claim should therefore be read with the domain stated: $x\in\mathbb{Z}$.

If the premise is instead understood to mean that $x+5$ is an odd integer, that condition itself implies $x$ is an integer, because subtracting the integer $5$ from an integer leaves an integer. The conclusion then follows as above.

Common mistakes

• Ignoring the domain: State that $x$ is an integer, or explain why the premise makes it an integer. Parity is not ordinarily assigned to arbitrary real numbers.
• Stopping after showing $x$ is even: The requested conclusion concerns $x^2$. Finish by writing $x=2m$ and showing $x^2=2(2m^2)$.
• Assuming the square is even without justification: Use the definition: a number is even when it equals $2$ times an integer. Here the integer is $2m^2$.
• Treating the implication as an equivalence: The proof establishes the stated direction, not its converse. For integer $x$, the converse is also true: if $x^2$ is even, then $x$ is even (an odd $x=2m+1$ would give $x^2=2(2m^2+2m)+1$, which is odd), so $x+5$ is odd.

A compact final proof is: since $x+5$ is odd, $x+5=2k+1$ for some integer $k$. Hence $x=2(k-2)$, so $x^2=2[2(k-2)^2]$, which is even.