$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers.

Question

$x$ is a rational number. Then, $\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. Any idea? Thanks!

I tried to solve this problem and I arrived to prove that in natural set (positive integers) we cannot have: $m^2 + n^2 = k^2$ and $m^2 + 2n^2 = l^2$. Also is rather easy to see that $m$ is odd, $n$ is even, also $n$ is multiple of $3$.

But I'm not sure that is the way.

Answer 1

Partial Answer

Let $x=\frac pq$ where GCD$(p,q)=1$. So, $x$ is in lowest possible terms.

Note if $p,q$ are two integers, then $p+q$, $p+2q$ and $p+3q$ will all leave different remainders upon division by $3$. Now note that, if $q\neq 3k$, GCD$(p+q,q)=$GCD$(p+2q,q)=$GCD$(p+3q,q)=1$, hence none of the fractions are further reducible.

Hence , there will always be one of the numerators of $3k+2$ form, since, as stated before, all leave different remainders. There is no perfect square of $3k+2$ form. Hence, no matter if $q$ is a perfect square or not, that particular number can never be rational.

If $q=3k$, $p=3m+2$ or $p=3m+1$. In the former case, the same argument applies, and a perfect square is impossible. However I'm still struggling with the latter case. Suggestions are welcome.

Answer 2

I tried to solve this problem and I arrived to prove that in natural set (positive integers) we cannot have: $m^2 + n^2 = k^2$ and $m^2 + 2n^2 = l^2$.

By definition, having $k^2-n^2 = m^2$ and $k^2+n^2=l^2$ would mean that $n^2$ is a congruum. But there exists no congruum that is a perfect square itself, by Fermat's right triangle theorem:

• A right triangle in the Euclidean plane for which all three side lengths are rational numbers cannot have an area that is the square of a rational number. The area of a rational-sided right triangle is called a congruent number, so no congruent number can be square.

  [...] equivalent to:

• If three square numbers form an arithmetic progression, then the gap between consecutive numbers in the progression (called a congruum) cannot itself be square.

Answer 3

Note that your question is equivalent to showing that for a rational $x$, all of $\sqrt{x-1}$, $\sqrt{x}$ and $\sqrt{x+1}$ cannot be rational. If possible, on the contrary, let that be true.

Let $x=\frac{a^2}{b^2}$ with $(a,b)=1$ so that \begin{align} \sqrt{x-1}&=\frac{\sqrt{a^2-b^2}}{b}\\ \sqrt{x}&=\frac{a}{b}\\ \sqrt{x+1}&=\frac{\sqrt{a^2+b^2}}{b} \end{align}

Since, $\sqrt{x+1}$ is rational, then $\sqrt{a^2+b^2}$ must be an integer. So, $$a^2+b^2=c^2$$ which implies \begin{align*} &a=p^2-q^2\\ &b=2pq\\ &c=p^2+q^2 \end{align*} for some integers $p,q,r$.

Similarly, $\sqrt{a^2-b^2}$ should also be an integer. So, $$(p^2-q^2)-4p^2q^2=x^2$$ for some integer $x$, which is not possible because of this.

This completes the proof.

Is that okay?