1. Irrational Number

Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.

2. Re: Irrational Number

Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
$\sqrt{1^3+3^3+3^3+4^3+5^1} = 2\sqrt{31} \in \mathbb{R} \setminus \mathbb{Q}$

... what now?

3. Re: Irrational Number

Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
Here are some comments. Is there a typo: $\sqrt{1^3+{\color{red}2^3}+3^3+4^3+5^n}~?$

Even if it is, nothing really changes in the question.

Also: if $k$ is a non-square positive integer then $\large\sqrt{k}$ is irrational.

4. Irrational number

Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.

6. Re: Irrational number

I still don't know how to solve it. Could you give me the complet explanation?

7. Re: Irrational number

in your other thread with the same problem, I just tried $n=1$ ... once again, on your edited expression I tried $n=1$ ...

$\sqrt{1^3+2^3+3^3+4^3+5^1} = \sqrt{105}$

As stated by Plato in your previous thread, note that 105 is not a perfect square, therefore its square root is irrational.

8. Re: Irrational Number

Yeah, there was a typo.

9. Re: Irrational number

Originally Posted by louis33
I still don't know how to solve it. Could you give me the complet explanation?
Originally Posted by louis33
Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
So the question reduces to: Find all $n\in\mathbb{N}^+$ such that $\Large5^n+100~$ is not a perfect square.

10. Re: Irrational Number

Let us find all $\displaystyle n$ for which $\displaystyle 5^n+100$ is a perfect square

$\displaystyle 5^n+100=m^2$ for some $\displaystyle m$

assume $\displaystyle n\geq 3$. Now m must be divisible by 5 so we can write m=5t.
replacing we get

$\displaystyle 5^{n-2}=(t-2)(t+2)$

so $\displaystyle t-2$ and $\displaystyle t+2$ are both powers of 5 but cannot both be divisible by 5 since otherwise their difference = 4 would be divisible by 5

therefore $\displaystyle t-2=1$ and $\displaystyle t+2=5^{n-2}$

this gives t=3 and n=3