Problem 1b

Find positive integers n such that $\sqrt { n-1 } +\sqrt { n+1 }$ is rational.

This a part of my set NMTC 2nd Level (Junior) held in 2014.

#NumberTheory #Integers

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Comments

With $n$ being a positive integer we would need to have both $n - 1$ and $n + 1$ as perfect squares. But since there is no pair of perfect squares that differ by $2$ we know that at least one of $\sqrt{n - 1}$ or $\sqrt{n + 1}$ will be irrational. Since both of these roots are non-negative, this implies that the sum will always be irrational.

A follow-up question could be: are there any real numbers $x \ge 1$ such that $\sqrt{x - 1} + \sqrt{x + 1}$ is rational?

On my first attempt, I've reduced the problem to the question of whether or not there are positive integers $a, b$ and $c$ such that $a^{4} - 3b^{4} = c^{2}$ . Haven't found any solutions yet, so I'm seeing if I can prove that there in fact can't be any solutions.

Brian Charlesworth - 6 years, 7 months ago

Since $\sqrt{ x -1 } + \sqrt{ x + 2 }$ is a continuous graph with image of $[\sqrt{2}, \infty )$ , hence there are many real values of $x$ which yield a rational value.

A slightly more interesting question is if there are rational solutions. My suspicion is no.

Calvin Lin Staff - 6 years, 7 months ago

@Calvin Lin Yes, that is the more interesting question. I think that there is a rational solution linked to every primitive Pythagorean triple $(a,b,c), a \lt b \lt c$ , such that (at least) one of $a,b$ is a perfect square, and, (letting, say, $b$ be the perfect square), that both $(c - b)$ and $(c + b)$ are perfect squares as well. For example, with $b = 4, c = 5$ we have

$\sqrt{\frac{5}{4} - 1} + \sqrt{\frac{5}{4} + 1} = \frac{1}{2} + \frac{3}{2} = 2$ .

Another example is the primitive triple $(36,77,85)$ . In this case we would have

$\sqrt{\frac{85}{36} - 1} + \sqrt{\frac{85}{36} + 1} = \frac{7}{6} + \frac{11}{6} = 3$ .

I suppose that for any Pythagorean triple there will be one of the legs $b$ such that both $c - b$ and $c + b$ are perfect squares, so what we require is that this leg $b$ is also a perfect square.

Looking through a list of primitives, other ones that would work are:

$(16,63,65)$ , giving a sum of $4$ ;
$(17,144,145)$ , giving a sum of $\frac{3}{2}$ ;
$(36,323,325)$ , giving a sum of $6$ ;
$(100, 621, 629)$ , giving a sum of $5$ .

Now that we know that such rational solutions exist, the next question is to determine how many there are, and if the resulting sums are integers or half-integers. I suspect that since there are an infinite number of primitive triples there will be an infinite number of rational solutions; I'll work on this later.

EDIT: O.k., with $c = m^{2} - n^{2}$ and $b = 2mn$ , where $m,n$ are positive integers with $m \gt n$ , we have

$\sqrt{\frac{c}{b} - 1} + \sqrt{\frac{c}{b} + 1} = \sqrt{\frac{2m}{n}}$ .

Then if $m = 2h^{2}$ and $n = k^{2}$ for positive integers $h,k$ we end up with a sum of $\frac{2h}{k}$ .

So it would appear that there are an infinite number of rational solutions. An unexpected result ......

Brian Charlesworth - 6 years, 7 months ago

It's never rational...

Nathan Ramesh - 6 years, 7 months ago

no integer satisfy n. The difference between the square of 2 consecutive integers must be atleast 3.

Ashutosh Kaul - 6 years, 7 months ago

There is no such integer

John Watson - 6 years, 7 months ago

Squaring it gives $2n+2\sqrt{n^{2}-1}$ which means that $n^{2}-1$ is a perfect square. Hence $n=1$ , but it does not work as $\sqrt {2}$ is irrational. Thus no solutions.

Joel Tan - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$