Proving the Solution a=4 and n=2 for a^(1/n) = √a; a>0, n>1

[TylerH]

$$\frac{a}{n}=\sqrt[n]{a};a > 0;n>1$$
How do I prove that a=4, and n=2 is the only solution? I'm pretty sure it is, but I don't know how to prove it.

 

[Preuton]

the only solution or the onl solution with natural numbers??

If is the only solution you could try to plot it to see how they behave

 

[Preuton]

a/n=b a^(1/n)=b

b*n = b^n

n=b^(n-1)

b=n ^(1/(n-1))

a= n^(n/(n-1))With this you can see that if n and a has to be natural there is only one sol and if they don't there are infinite ones.

 

[TylerH]

You were correct in assuming that I meant integers. I also plotted it. The shape is interesting, it almost appears linear. Thanks for your help, much appreciated.

 

[TylerH]

Wouldn't that result in a dependency? How can you say n^(n/(n-1)) has only one integer solution?

 

[Preuton]

n^(n/(n-1))=n^(1+1/(n-1))=n+n^(1/(n-1))

for integers values of n

n= integrer

a=n^(1/(n-1))= only integrer for n=2;

dem: function is decreasing , for n=2 , a=2 for n -> inf a=1

 

[TylerH]

Oh... that was a dumb question... Thanks. :)

 

[TylerH]

Here's why I was asking:
(aside: how do I impose restrictions on the possible values of a and b? Like say that they must not be equal and that they must be integers?)
$$a^b=b^a$$
Which implies that b must equal a, by a factor, n, and a must equal b by a power of 1/n. To make it easier, I'll switch the n and 1/n, since the actual value doesn't matter, just the relation. We can also assume a must be a power of b, because otherwise there would be extraneous factors, making it impossible for them to be equal. Thus we have:
$$\frac{b}{n}=a;a^n=b$$
Which you can tell is very similar to the question posed, except the original was simplified.
$$an=a^n$$
$$n=a^{n-1}$$
$$a=\sqrt[n-1]{n}$$