Find all positive integers n

anemone

MHB POTW Director
Staff member
Find all positive integers $$\displaystyle n$$ for which $$\displaystyle \sqrt{n+\sqrt{1996}}$$ exceeds $$\displaystyle \sqrt{n-1}$$ by an integer.

chisigma

Well-known member
Find all positive integers $$\displaystyle n$$ for which $$\displaystyle \sqrt{n+\sqrt{1996}}$$ exceeds $$\displaystyle \sqrt{n-1}$$ by an integer.
The condition leads to the equation...

$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$

... and squaring both terms of (2) we arrive to write...

$$2\ \sqrt{499} = k^{2} + 2\ k\ \sqrt{n-1} - 1\ (2)$$

The only solution of (2) for n and k integers seems to be n=500 and k=1...

Kind regards

$\chi$ $\sigma$

topsquark

Well-known member
MHB Math Helper
The only solution of (2) for n and k integers seems to be n=500 and k=1...
Thank Heavens! My head was insisting there be more solutions, otherwise it seemed too easy.

Of course now I just jinxed myself!

-Dan

Jester

Well-known member
MHB Math Helper
First

$\displaystyle \lim_{n \to \infty} \sqrt{n+\sqrt{1996}}-\sqrt{n-1} = 0$.

It not hard to show that the function is decreasing and when $n = 1$, $\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=6.758447736$ then so you really only need to check when the difference is either $1, 2, 3, 4, 5$ or $6$ and only 1 gives that $n$ is an integer.

anemone

MHB POTW Director
Staff member
Thanks to those for participating in this challenge problem.

My solution:

Let $$\displaystyle \sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k$$, where $$\displaystyle k$$ is an integer.

$$\displaystyle \sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}$$

$$\displaystyle n+\sqrt{1996}=k^2+n-1+2k\sqrt{n-1}$$

$$\displaystyle n+\sqrt{1996}=k^2+n-1+\sqrt{4k^2(n-1)}$$

By equating the radicand from both sides, we get

$$\displaystyle 1996=4k^2(n-1)$$

$$\displaystyle 499=k^2(n-1)$$

Since 499 is a prime number, and $$\displaystyle k^2=1$$, we can conclude that

$$\displaystyle n-1=499$$ or $$\displaystyle n=500$$ is the only solution to the problem.

topsquark

MHB Math Helper

-Dan

Well-known member
The condition leads to the equation...

$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$

... and squaring both terms of (2) we arrive to write...

$$2\ \sqrt{499} = k^{2} + 2\ k\ \sqrt{n-1} - 1\ (2)$$

The only solution of (2) for n and k integers seems to be n=500 and k=1...

Kind regards

$\chi$ $\sigma$
nice deduction by making radicals same on both sides