## KingGeorge 3 years ago [SOLVED] KingGeorge's Challenge of the Month! (this might be easier than previous challenges) Prove that $$\sqrt{n+\sqrt[3]{n+1}}$$ is irrational for all $$n\in\mathbb{N}$$.

1. ParthKohli

2. KingGeorge

If I wrote it formally, I believe I would use a proof by contradiction.

3. ParthKohli

$\Rightarrow\sqrt{n + \sqrt[3]{n + 1}} = {p \over q}$p and q are co-prime. Square both sides.$\Rightarrow n + \sqrt[3]{n + 1} = {p^2 \over q^2}$$\Rightarrow nq^2 + q^2\sqrt[3]{n + 1} = p$I failed.

4. ParthKohli

Oops. p^2 in the last reply*

5. ParthKohli

I am not the best at classic proofs; this was my attempt(and I don't even know if it was correct).

6. KingGeorge

Not the way I would do it.

7. ParthKohli

*gives up*. Good luck, all of you geniuses who'd solve this problem. Cheers!

8. Hero

negative

9. ganeshie8

using @dominusscholae method of rational root test

10. ganeshie8
11. ganeshie8

$$x = \sqrt{n+\sqrt[3]{n+1}}$$ $$x^2 = n+\sqrt[3]{n+1}$$ $$x^2-n = \sqrt[3]{n+1}$$ $$(x^2-n)^3 = n+1$$ $$x^6-3x^4n + 3x^2n^2 - n^3 = n+1$$ $$x^6-3nx^4 + 3n^2x^2 - (n^3+n+1)$$

12. ganeshie8

this is a polynomial in x with integer coefficients

13. ganeshie8

if this polynomial has any rational roots at all, they should be of form x = $$\pm(n^3+n+1/1), \pm(1/1)$$ 1) put x =1, n = 1 in the polynomial 2) put n=1, x= 3 in the polynomial none of them satisfy the polynomial. and hence x is not rational

14. dominusscholae

^Right idea, perhaps wrong explanation. You can say that $\pm(n^3 + n + 1)$ has no factors, and then plug it into the polynomial constructed. I'll say this: This method is extremely algebra intensive.....

15. dominusscholae

I'd probably say that mukushla's involves lesser steps.

16. Neemo

I think there is a problem with the question ! Since $\sqrt{n+\sqrt[3]{n+1}}$ is an integer for n= 0 and n=7

17. dominusscholae

Lol. Perhaps a small definitional error. Still, if this is the case, the proof wouldn't matter. Maybe a rewording of the question saying all numbers of the form above are irrational :P.

18. dominusscholae

WHOOPS! Did I say the proof wouldn't matter?!? XD. I think I meant to say that the fact that only n = 0 to n = 7 would make this valid wouldn't matter.

19. mukushla

@Neemo....good catch...lol.... Correction let $$\frac{a}{b}=\sqrt{n+\sqrt[3]{n+1}}$$ $$a$$ , $$b$$ are positive integers and $$\gcd(a,b)=1$$ . some algebra gives $$b^6 n^3-3a^2b^4n^2+(b^6+3b^2a^4)n+b^6-a^6=0$$ easily u can see $$\gcd(b^6-a^6,b^6)=1$$ since $$\gcd(a,b)=1$$. if $$b \neq 1$$ ....according to rational root theorem there is no natural root for $$n$$. because $$\gcd(b^6-a^6,b^6)=1$$...this is a contradiction... so its better say if $$\sqrt{n+\sqrt[3]{n+1}}$$ is not an integer then it will be irrational...

20. Neemo

that's what I was thinking ! from the start @mukushla with same @mukushla's notations(and algebra too) $\exists m \in \mathbb{N}-----bm=a^6$ then b divides a^6 But, gcd(a,b)=1=gcd(a^6,b) then b=1 so if $\sqrt{n+\sqrt[3]{n+1}} \text{ is rational ,then It's an integer }$

21. mukushla

$$\text{or we can say if} \ \sqrt{n+\sqrt[3]{n+1}} \ \text{is not integer then it will be irrational...}$$

22. KingGeorge

Thanks for pointing out 7 as a solution @neemo. I was not including 0 as part of the natural numbers which is why I wasn't worried about that. To make the rewording, show that 7 is the only rational number of the form $$\sqrt{n+\sqrt[3]{n+1}}$$ where $$n\in\mathbb{N}$$.

23. mukushla

actually $$n=2196$$ makes it integer too...im wondering how we can find all solutions for which that expression becomes an integer

24. KingGeorge

I guess I need to be more careful when writing these problems then. Good job for breaking the problem.

25. mukushla

how about the proof...are u agree with our work...and the conclusion ''...is not integer then it will be irrational...''

26. KingGeorge

I agree with your work and conclusion. Once again, good job.

27. Hero

Impressive Stuff @KingGeorge, @mukushla, @Neemo

28. mukushla

@KingGeorge thank u.............:) and hero too

29. KingGeorge

Now I need to come up with another problem of the month that is actually written correctly :/

30. mukushla

:)

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