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Can anyone help me with an analytical proof for this? This was a question on a Data Sufficiency question; and while I can empirically see the answer, proving it analytically is beyond my abilities.

The Short One
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Abhishek Chakrabarti
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  • Because $~\sqrt[\large^n]2~\not\in~\mathbb N~$ for $~n>1$. – Lucian Mar 19 '15 at 09:05
  • I think what Abhishek means is: Let $x$ be a real number, and suppose that $x^2$ and $x^3$ are both even integers. Why does it follow that $x$ is an even integer? – TonyK Mar 19 '15 at 22:24

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If $x^2$ is even and $x$ is a whole number, then $x$ is even. This is easiest to prove by proving that if $x$ is odd, then $x^2$ is odd.

5xum
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  • Thank you. However, as the question stands, I can't assume that x is an integer/natural number at the start. Thus, by combining x squared and x cubed being even; I somehow restrict x to the set of even numbers. – Abhishek Chakrabarti Mar 19 '15 at 15:58
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    @AbhishekChakrabarti Well, if $x^2$ is even, then it is a natural number, meaning $x=\sqrt{n}$ for some $n$, so $x^3=n\sqrt{n}$. From this, you can quickly show that $x$ must be a whole number. – 5xum Mar 19 '15 at 16:44
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I don't know if this is analytical enough for you, but here goes: let's say $x = 2n$. Then $x^2 = 2^2 n^2$ and $x^3 = 2^3 n^3$. But if $x = 2n + 1$, then $x^2 = 4n^2 + 4n + 1$ and $x^3 = 8n^3 + 12n^2 + 6n + 1$.

The Short One
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Since $x^2$ and $x^3$ are even by assumption, there are natural numbers $n,m\in \mathbb N$ such that $$ x^2 = 2n, x^3 = 2m. $$ Trying to produce $x$ by division gives $$\begin{split} x &= \frac{x^3}{x^2} =\frac{2m}{2n} = \frac mn \\ &= \frac{x^4}{x^3} = \frac{4n^2}{2m} = \frac{2n^2}{m}, \end{split}$$ which implies $$ m^2 = 2n^3. $$ Hence $m$ must be divisible by $n$, so $x$ is a natural number. Since $x^2$ is even, $x$ cannot be odd.

daw
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If $x^2\in2\mathbb{Z}$ and $x^3\in2\mathbb{Z}$, then $x=x^3/x^2\in\mathbb{Q}$.

Since $x^2-2k=0$ for some $k\in\mathbb{Z}$, we have that $x$ is an algebraic integer.

As shown in this answer, the intersection of $\mathbb{Q}$ and the algebraic integers is $\mathbb{Z}$. Thus, $x\in\mathbb{Z}$. Now we can use the fact that $x$ must be even, for if it were odd, $x^2$ would be odd.

Community
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robjohn
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The question might be better phrased as

If $x^2$ and $x^3$ are both even integers, is $x$ necessarily an even integer?

I don't think the question is a priori assuming that $x$ is even an integer.

I would begin by showing that since $x^2$ and $x^3$ are both integers, $x=x^3/x^2$ must be a rational number.

A slightly challenging (but very standard) thing to prove here is that if the square root of an integer is rational, then it is also an integer. Hence $x=\pm\sqrt{x^2}$ is an integer (since it is the rational square root of an integer.)

Since $x$ is an integer, it is either even or odd. If it is odd, then $x^2$ and $x^3$ are both odd. Thus $x$ must be even, since $x^2$ and $x^3$ are certainly not both odd.

Justin Lindberg
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First $x=\dfrac{x^3}{x^2}$ is a rational number. Write $x=\dfrac mn$, $\,m\wedge n=1$. Then $x^2=\dfrac{m^2}{n^2}$ is an (even) integer. As $m^2$ and $n^2$ are coprime, this implies $n^2=1$, hence $x$ in an integer.

Now, if $x$ is an integer and $x^2$ is even, $x$ is even (by contraposition: if $x$ were odd, $x^2$ would be too).

Bernard
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