For which natural numbers $n ≥ 999$ is the number $N = \sqrt{n-999}+\sqrt{n+1000}$ natural?

Question

my question

For which natural numbers $n ≥ 999$ is the number $N = \sqrt{n-999}+\sqrt{n+1000}$ natural?

my idea

I tried all the variants but it always gives that $n=999*1000$, which wont get $a$ and $b$ to be perfecr numbers

Answer 1

You can solve that in an analytics way. We know that $n-999$ and $n+1000$ must be squares. So we have $$ n - 999 = p^2 \text { for some } p \in\mathbb Z $$ and $$ n +1000 = q^2 \text { for some } q \in\mathbb Z. $$ Let us write $n = p^2 + 999$ and use it in the other equation: $$ p^2 + 999 + 1000 = q^2 \Leftrightarrow p^2 + 1999 = q^2. $$ Now we can write $ 1999 = q^2-p^2 $ and that means, we are looking for the difference of two squares. But since $1999$ is odd, there exists at least one. And since it is even a prime, there exists exactly one, namely $$ 1999 = 1999\cdot 1 = (\frac{1999+1}{2})^2 - (\frac{1999-1}{2})^2 = 1000^2 - 999^2. $$ So we have $q = 1000$ and $p = 999$ which leads to $$ n = 999^2 + 999 = 999000. $$ Indeed, since it is also $$ n = 1000^2 - 1000 = 999000. $$

And since there are no other solutions for the difference of two squares, this is the only solution for your question.

Answer 2

$n=999000$ is a solution $$\sqrt{999000+1000}+\sqrt{999000-999}$$ $$\sqrt{1000000}+\sqrt{998001}$$ $$1000+999$$ $$1999$$

I did what you did.