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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.

Note by Siddharth G
2 years, 11 months ago

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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 - 2 years, 11 months ago

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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 - 2 years, 11 months ago

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@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 - 2 years, 11 months ago

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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 - 2 years, 11 months ago

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There is no such integer

John Watson - 2 years, 11 months ago

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

Ashutosh Kaul - 2 years, 11 months ago

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It's never rational...

Nathan Ramesh - 2 years, 11 months ago

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