Does this wierd way work, as it is very different from Kalva's answer

[/hide]

I get this strange feeling I'm missing something, since this seemed really easy... Oh well, I'll post my solution anyway.

Mine solution is: a^n=a+1 <=> 1) a^2n=a^2+2a+1. 2) b^2n=b+3a 1)-2) gets: a^2n-b^2n=a^2+1-(a-b) a^2n-b^2n+(a-b)=a^2+1 (a-b)(a+b)(a^2+a^2)(a^4+b^4)...(a^n+b^n)+(a-b)=a^2+1 (a-b)((a+b)(a^2+b^2)(a^4+b^4)...(a^n+b^n)+1)=a^2+1 <=> (a,b - positive) (a-b)(POSITIVE NUMBER)=(POSITIVE NUMBER) .... so a-b>0 a>b => a is larger than b.

We square and rearrange the first equation and also rearrange the second. \[ a^{2n}-a=a^2+a+1\] \[ b^{2n}-b=3a\] It is trivial that $a>1$, $b>1$, thus \[ a^2+a+1>3a\] \[ a^{2n}-a>b^{2n}-b\] where we substituted to achieve the result. This writes $(a-b)(a^{2n-1} + a^{2n-2}b + \cdots + ab^{2n-2} + b^{2n-1} - 1) > 0$, and as the second factor is positive, it follows $a>b$.

Posted before at least 3 times. \[{ a-b=\frac{(a-1)^2}{a^{2n-1}+a^{2n-2}b+...+b^{2n-1}-1}}>0.\]

@limac, I know this is a really late post, but you can show that: $f(x)=x^(2n)-x$ is an increasing function when x>1. Wouldn't that do the trick. Are there any other ways of proving this?

Let assume for sake of contradiction that $b>a$, and as above $a,b>1$ so, $b^{2n}>a^{2n}$, $b+3a>a^2+2a+1$, $b+a>a^2+1$, since we assume that $b>a$, $b/a>1$, so $2>b/a+1>a+1/a$, so $2>a+1/a$, contradiction, since $a>1$

I claim that $a>b$ for all positive integers $n$. First note that from the equality $a^n=a+1$ we have that raising $a$ to an integer power increases its value, implying that $a>1$. Similarly, we may deduce $b>1$. Now note that we can write $a^{2n}=a^2+2a+1$, so\[a^{2n}-b^{2n}=(a^2+2a+1)-(b+3a)=(a^2-a+1)-b.\]In order to prove our claim, it suffices to show that $a^2-a+1>b$ for all $n$. If we can do this, then we have $a^{2n}-b^{2n}>0$, so $a^{2n}>b^{2n}\implies a>b$. Suppose the contrary, that $b\geq a^2-a+1$. As $b$ is positive, increasing $b$ increases the discrepancy between $b^{2n}$ and $b$. Therefore, if we can show that $b^{2n}>b+3a$ for only the case $b=a^2-a+1$, then our desired follows. Note that from $a>1$ we have $a-1>0$, and squaring this gives $a^2-2a+1>0\implies a^2-a+1>a$. Now raising both sides to the $(2n)^{\text{th}}$ power gives\[(a^2-a+1)^{2n}>(a^n)^2=(a+1)^2=(a^2-a+1)+3a\implies b^{2n}>b+3a,\] contradiction. Therefore $a^2-a+1>b$ for all positive integers $n$, and we are done. $\blacksquare$

Note that $a^n=a+1>1\Rightarrow a>1$ and, thus, $b^{2n}=b+3a>3\Rightarrow b>1$. Now noting that $(a-1)^2>0\Rightarrow a^2+1>2a$, \begin{align*} a^{2n}-a&=a^2+a+1\\ &>3a\\ &=b^{2n}-b \end{align*} Now, note that $f(x)=x^{2n}-x$ is increasing for all $x>1$ (seen easily by differentiating). Thus, we may conclude that $f(a)>f(b)\Rightarrow a>b$.$\blacksquare$

Another solution using Calculus: First its easy to notice using the expressions that $a>1,b>1$ Consider the function $f(x)=x^{2n}-x-3a$ for $x\in [1,\infty)$ clearly $x=b$ is a root of it. We see that $f'(x)>0$ for all $x\in [1,\infty)$, thus $f$ is strictly increasing on this interval. We know that such a function can intersect $x-$axis at most once thus $b$ is the only possible root. Now we have $f(a)=(a-1)^2>0$ and $f(1)<0$ thus using Intermediate Value Theorem $1<b<a$.