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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, 4 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. · 2 years, 4 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. Staff · 2 years, 4 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 ...... · 2 years, 4 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. · 2 years, 4 months ago

There is no such integer · 2 years, 4 months ago

no integer satisfy n. The difference between the square of 2 consecutive integers must be atleast 3. · 2 years, 4 months ago