# An ugly formula [closed]

Our teacher wrote the following ugly algebraic formula on the blackboard $$(n+\sqrt{n^2-1})^{4/3}+(n+\sqrt{n^2-1})^{-4/3}$$ Our teacher told us that for some positive integers $n$ this ugly formula has integer value! Then our teacher asked us: "What is the largest integer value below 1 million that can be produced by putting a positive integers $n$ in this formula?" Now I ask you: Can you find the answer?

## closed as off-topic by f'', The Dark Truth, Engineer Toast, manshu, CodeNewbieJul 29 '16 at 8:27

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• @Haobin computers are not allowed? – user230452 Mar 20 '16 at 16:04
• The tag no-computers is A puzzle designed to be solved without using calculators or computers. – Haobin Mar 20 '16 at 16:14
• Personally, I'd guess that the answer is 2, obtained when n=1 and the teacher is laughing right now. – Hugh Meyers Mar 20 '16 at 17:11
• >! $x=n+\sqrt{n^2-1}$ is a solution to $x+1/x=2n$ – Johannes Mar 20 '16 at 18:26

@Tony Ruth's answer provides this alternate form of the equation:

$16n^4 - 16n^2 + 2 = m^3 -3m$

Adding 2 to both sides and factoring,

$(4n^2-2)^2=(m-1)^2(m+2)$

Then

$m+2=(\frac{4n^2-2}{m-1})^2=x^2$ (with $x>0$, and $x$ integer)

Substituting this back in,

$x^2(x^2-3)^2=(4n^2-2)^2$

So

$x(x^2-3)=4n^2-2$ (both sides are positive)

Add 2 to both sides again:

$(x-1)^2(x+2)=(2n)^2$

So

$x+2=(\frac{2n}{x-1})^2=y^2$ (with $y>0$, and $y$ integer)

And

$y(y^2-3)=2n$ (again, with both sides positive)

For every $y>1$, this is true for some integer $n$, so we can choose the $y$ that gives the biggest value of $m$ less than 1 million. This is:

$y=\lfloor\sqrt[4]{1,000,000}\rfloor=31$
$m=(y^2-2)^2-2=919,679$
$n=\frac{y(y^2-3)}{2}=14,849$.

$919,679$

Partial solution. I can reduce this to a polynomial equation with only integral powers, but so far I am not sure how to find the integer solutions of this new equation.

New equation:

$16n^4 - 16n^2 + 2 = m^3 -3m$

where m < 1,000,000

This is a depressed cubic equation for m, I tried setting $16n^4 - 16n^2 + 2=q$ and solving the depressed cubic, and I get a new ugly equation $m = (\frac{q + \sqrt{q^2-4}}{2})^\frac{1}{3} + (\frac{2}{q + \sqrt{q^2-4}})^{\frac{1}{3}}$

Here is how I derived the new equation.

Square the original equation $$(n+\sqrt{n^2-1})^\frac{8}{3} + (n+\sqrt{n^2-1})^\frac{-8}{3} + 2 = m^2$$

Then

Move the 2 over and multiple this equation by the original equation. $$(n+\sqrt{n^2-1})^4 + (n+\sqrt{n^2-1})^{-4} + (n+\sqrt{n^2-1})^\frac{4}{3} + (n+\sqrt{n^2-1})^\frac{-4}{3} = m^3 - 2m$$

Substitute

$$(n+\sqrt{n^2-1})^4 + (n+\sqrt{n^2-1})^{-4} = m^3 - 3m$$

Multiply by $$(\frac{n-\sqrt{n^2-1}}{n-\sqrt{n^2-1}})^4$$ and create a like denominator

You get,

$$(\frac{(n+\sqrt{n^2-1})*(n^2-(n^2-1))}{n^2-(n^2-1)})^4+(\frac{(n-\sqrt{n^2-1})}{n^2-(n^2-1)})^4$$

Then

The $n^2-(n^2-1)$ terms are just 1. Finally you expand out the powers and all the terms with radicals cancel.

• If you say that (4n²-1)²'s a constant, then you get an equation of the type m³ - pm = q, from there you could try solving the depressed cubic – Chill Fruit Mar 20 '16 at 16:56
• @Gamow, Thanks, I had made an algebra error. I updated the equation. – Tony Ruth Mar 20 '16 at 19:51
• Can you explain how you factorised it to get the new equation? – user230452 Mar 20 '16 at 22:58
• @user230452 I added the derivation. It's fairly straightforward. Finding $m^3-3m$ removes the $\frac{4}{3}$ power. Then conjugate to get rid of the radical. – Tony Ruth Mar 21 '16 at 0:52