Find all $n\in \mathbb N$ such that $\sqrt{n}+\sqrt{n+m}\in\mathbb{N}$, where $m$ is the product of two primes that are not equal to each other.

Question

I have the following problem:

Find all $n$ that are natural numbers such that $\sqrt{n}+\sqrt{n+m}\in\mathbb{N}$, where $m$ is a positive natural number that is the product of two primes that are not equal to each other.

I thought that I can solve it when the function under the square roots are squares of integers but I have no idea how to prove that or how to continue.

Answer 1

Assignment:

Find all $\text{n}\in\mathbb{N}$ such that $\sqrt{\text{n}}+\sqrt{\text{n}+\text{m}}\in\mathbb{N}$, where $\text{m}\in\mathbb{N}^+$ and $\text{m}=\text{p}_1\cdot\text{p}_2$ where $\text{p}_1\wedge\text{p}_2\in\mathbb{P}$ and $\text{p}_1<\text{p}_2$.

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Solution:

Notice that when $\text{n}$ and $\text{n}+\text{m}$ are both perfect squares, the requirements are met.

So, let $\text{n}=\text{y}^2$ and $\text{n}+\text{m}=x^2$. Than we know that:

$$x^2-\text{y}^2=\left(x+\text{y}\right)\left(x-\text{y}\right)=\text{n}+\text{m}-\text{n}=\text{m}\tag1$$

Now, let's look at a few cases:

• Case 1:

Write $\left(x+\text{y}\right)\left(x-\text{y}\right)=\text{m}$ as: $\left(x+\text{y}\right)\left(x-\text{y}\right)=\text{m}\cdot 1$. Because we know that $x$, $\text{y}$ and $\text{m}$ are positive integers we also know that:

$$ \begin{cases} x+\text{y}=\text{m}\\ \\ x-\text{y}=1 \end{cases}\space\space\space\Longleftrightarrow\space\space\space\begin{cases} x=\frac{\text{m}+1}{2}\\ \\ \text{y}=\frac{\text{m}-1}{2} \end{cases}\tag2 $$

This implies that:

$$\text{n}=\text{y}^2=\left(\frac{\text{m}-1}{2}\right)^2\tag3$$

So, we get:

$$\sqrt{\text{n}}+\sqrt{\text{n}+\text{m}}=\sqrt{\left(\frac{\text{m}-1}{2}\right)^2}+\sqrt{\left(\frac{\text{m}-1}{2}\right)^2+\text{m}}=$$ $$\sqrt{\left(\frac{\text{m}-1}{2}\right)^2}+\sqrt{\left(\frac{\text{m}+1}{2}\right)^2}=\frac{\text{m}-1}{2}+\frac{\text{m}+1}{2}=\text{m}\tag4$$

• Case 2:

Write $\left(x+\text{y}\right)\left(x-\text{y}\right)=\text{m}$ as: $\left(x+\text{y}\right)\left(x-\text{y}\right)=\text{p}_1\cdot\text{p}_2$. Because we know that $x$, $\text{y}$ and $\text{m}$ are positive integers we also know that:

$$ \begin{cases} x+\text{y}=\text{p}_2\\ \\ x-\text{y}=\text{p}_1 \end{cases}\space\space\space\Longleftrightarrow\space\space\space\begin{cases} x=\frac{\text{p}_1+\text{p}_2}{2}\\ \\ \text{y}=\frac{\text{p}_2-\text{p}_1}{2} \end{cases}\tag5 $$

This implies that:

$$\text{n}=\text{y}^2=\left(\frac{\text{p}_2-\text{p}_1}{2}\right)^2\tag6$$

So, we get:

$$\sqrt{\text{n}}+\sqrt{\text{n}+\text{m}}=\sqrt{\left(\frac{\text{p}_2-\text{p}_1}{2}\right)^2}+\sqrt{\left(\frac{\text{p}_2-\text{p}_1}{2}\right)^2+\text{p}_1\cdot\text{p}_2}=$$ $$\sqrt{\left(\frac{\text{p}_2-\text{p}_1}{2}\right)^2}+\sqrt{\left(\frac{\text{p}_1+\text{p}_2}{2}\right)^2}=\frac{\text{p}_2-\text{p}_1}{2}+\frac{\text{p}_1+\text{p}_2}{2}=\text{p}_2\tag7$$

Now, I let you prove if these are the only solutions. HINT: write $\text{k}=\sqrt{\text{n}}+\sqrt{\text{n}+\text{m}}$ and solve for $\text{n}$, which gives:

$$\text{n}=\frac{1}{4}\left(\text{k}-\frac{\text{m}}{\text{k}}\right)^2\tag8$$

Answer 2

Let $$\sqrt{n}+\sqrt{n+m}=k\in\mathbb{N}$$ then

$$2\sqrt{n(n+m)}=\underbrace{k^2-n-m}_{a\in\mathbb{N}} \;\;\;/^2$$

So $$4n^2+4nm = a^2+m$$ and thus $$\underbrace{4n^2+4nm+m^2}_{(2n+m)^2} = a^2+m^2$$

If $m=pq$ then $$(2n+m)^2-a^2 = p^2q^2$$

or $$\boxed{(2n+m-a)(2n+m+a)=p^2q^2}$$

Now we have same cases which I'm not going to do it all.

• If $2n+m+a= p^2q^2$ and $2n+m-a=1$, so $$4n+2pq=p^2q^2+1$$ so $$ n = {(m-1)^2\over 4}$$
• If $2n+m+a= p^2q$ and $2n+m-a=q$, so $$4n+2pq=p^2q+q$$

so $$ n = {q(p-1)^2\over 4}$$ (Note: You have to check if this actually works.)

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