0
$\begingroup$

Find all $n\in \mathbb N$ such that $\sqrt{n+7}+\sqrt{n}$ is rational.

By inspection it is pretty easy to see that the only $n$ that will work is $n=9$. Because the distance between perfect squares grows infinitely bigger, after $n=9$ we will never have another number that you can add seven to and get a perfect square. I.e.

$25-7=18\quad$ (Need $16$, difference of $2$)
$36-7=29\quad$ (Need $25$, difference of $4$)
$49-7=42\quad$ (Need $36$, difference of $6$)
$64-7=57\quad$ (Need $49$, difference of $8$)

So on and so forth. It is easy to see we will never have another $n\in \mathbb N$ that will work but I am having trouble coming up with a sophisticated way to explain this.

I'm trying to find a way that uses something like:

For $a,b\in \mathbb Z, \sqrt{\frac{a}{b}}$ is rational if and only if $\sqrt{ab}$ is rational. And, for $m\in \mathbb Z, \sqrt{m}$ is rational if and only if $\sqrt{m}\in\mathbb N$.

I'm having trouble making the connection between knowing $n=9$ and the part with $a,b,$ and $m$.

Any suggestions are appreciated.

$\endgroup$
  • $\begingroup$ We also have that for nonnegative integers $a, b$, $\sqrt{a} + \sqrt{b}$ is rational iff $a, b$ are perfect squares. This (which you might like to prove yourself) combined with your observation does the job. $\endgroup$ – Travis Willse Sep 24 '14 at 5:39
2
$\begingroup$

Once you've shown that for positive integers $x,y$, $\sqrt{x}+\sqrt{y}$ is rational iff $x,y$ are perfect squares, the following will help you show that the only perfect squares whose difference is $7$ are $16$ and $9$.

If $n+7$ and $n$ are both perfect squares, then, $n+7 = a^2$ and $n = b^2$ for positive integers $a,b$.

Then, we have $(a-b)(a+b) = a^2-b^2 = (n+7)-n = 7$.

So, $a-b$ and $a+b$ are two integers whose product is $7$. How many integer factors does $7$ have?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Well, the only positive integers whose product is $7$ are $1$ and $7$. Then, either $(a-b)= 1$ or $(a-b)=7$ and the same goes for $(a+b)$. But, because $a,b \in \mathbb Z$, $(a+b)\ne 1$. So, $(a+b)=7$ and $(a-b)=1$. Then by substitution, $b=3$ which yields $a=4$ and $n=9$. $\endgroup$ – Vincent Sep 24 '14 at 6:01
1
$\begingroup$

$$\sqrt{n+7}=a-\sqrt n\implies n+7=a^2+n-2a\sqrt n\iff\sqrt n=\frac{a^2-7}{2a}$$

$$\implies n=\left(\frac{a^2-7}{2a}\right)^2 $$ where $a$ is rational

$$n+7=\frac{(a^2-7)^2}{4a^2}+7=\frac{(a^2+7)^2}{4a^2}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If I am not wrong didn't you missed an n. $\endgroup$ – Jasser Sep 24 '14 at 5:56
  • $\begingroup$ @user291957, Please pinpoint the mistake $\endgroup$ – lab bhattacharjee Sep 24 '14 at 6:02
  • $\begingroup$ Apologies for my bad observation that is all great. $\endgroup$ – Jasser Sep 24 '14 at 6:06
0
$\begingroup$

Hint : $\sqrt(n+7) +\sqrt(n)=a$ where a is an rational number. Now square on both sides and then simplify for n by again squaring and see the result.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.