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Number Theory Level 3

Take some time to read the following statements.

$[1]$ It is impossible to write $1$ as the sum of the reciprocals of $n$ odd integers when $n$ is an even number.

$[2]$ If $n$ is an integer greater than $11$ , then it is always possible to write $n$ as the sum of two composite numbers.

$[3]$ If $a$ is an irrational number, it is impossible for $a^{\sqrt{2}}$ to be a rational number.

Which of these statements are correct?

Note : This problem is a part of the set "I Don't Have a Good Name For This Yet". See the rest of the problems here . And when I say I don't have a good name for this yet, I mean it. If you like problems like these and have a cool name for this set, feel free to comment here .

All of them are correct

[1]

and

[2]

[1]

and

[3]

Only

[1]

1 solution

Sean Ty
Jul 30, 2014

$1.$ We first assume that 1 can be written as the sum of the reciprocals of $n$ odd integers when $n$ is an even number.

Let $\displaystyle \dfrac{1}{a_1}+\dfrac{1}{a_2}+....+\dfrac{1}{a_n}=1$ (Where $a_1,a_2,....a_n$ are odd numbers and $n$ is even.)

Taking their sum, we get $\displaystyle \dfrac{\dfrac{m}{a_1}+\dfrac{m}{a_2}+.....+\dfrac{m}{a_n}}{m}$ (Where $\displaystyle m=\prod_{i=1}^{n}a_i$ )

Since $a_1, a_2, ....a_n$ are all odd, then their product $m$ is also odd. By parity, the expression $\displaystyle \dfrac{\dfrac{m}{a_1}+\dfrac{m}{a_2}+.....+\dfrac{m}{a_n}}{m}$ is even. And so it is impossible to obtain a $1$ .

$2.$ Let $a>11$ .

We can rewrite $a=(a-4)+4=(a-6)+6=(a-8)+8$ .

One of $a-4,a-6,a-8$ is divisible by 3.

So it is always possible to write $n$ as the sum of $2$ composite numbers.

$3.$ I merely used the case of $e^{i\pi}=-1$ . So pretty much a guess. Anyone has a proof for this?

$(\sqrt{2}^\sqrt{2})^{\sqrt{2}}=2$

lawrence Bush - 6 years, 10 months ago

I misread the third statement and thought it said it is POSSIBLE, not IMPOSSIBLE. Ugh.

For #3, I don't see what you mean by the $e^{i \pi} = -1$ example. Care to explain what you mean? The most commonly known way to solve #3 is this:

Let $P = \sqrt{2}^{\sqrt{2}}$ . Then if $P$ is rational, we are done ( $a = \sqrt{2}$ ). If not, then $P$ is irrational, and $P^{\sqrt{2}} = (\sqrt{2})^2 = 2$ and this is rational. Thus, the statement is proven.

Though unimportant to the proof, it has been proven that $P$ is irrational.

Michael Tong - 6 years, 10 months ago

@Sean Ty , I don't understand your argument for 3.

Mursalin Habib - 6 years, 9 months ago

Haha, I'm sorry. It was meant to be a troll. Actually, from the choices I could already pick my answer. Sorry for the misconceptions!

Sean Ty - 6 years, 9 months ago

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