Proving that there exists $n,a \in \mathbb{N}$ such that $n = \sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+...}}}}$ for all $n \in \mathbb{N}$ I suspect that there are infinite solutions of $n,a \in \mathbb{N}$ for all $n \in \mathbb{N}$, $n>1$ such that $$n = \sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+...}}}}$$
Some solutions include, for $(n,a):$ $(2,2), (3,6), (4,12), (5, 30), (6,42), (7,56)...$. However, I wish to prove this and I'm not sure how/if my proof is right. This is my work:
$$n=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+...}}}}$$
$$n = \sqrt{a+n}$$
$$n^2-n=a$$
$$n^2-n+\frac{1}{4}=a+\frac{1}{4}$$
$$(n-\frac{1}{2})^2=a+\frac{1}{4}$$
$$n= \frac{1}{2} \pm \sqrt{a+\frac{1}{4}}$$
Since we are looking for values in $\mathbb{N}$ (positive solutions),
$$n=\frac{1+\sqrt{4a-1}}{2}$$
We can also find $a$ in terms of $n$ by evaluating
$$n=\frac{1+\sqrt{4a-1}}{2}$$
$$2n-1=\sqrt{4a+1}$$
$$\frac{(2n-1)^2-1}{4}=a$$
Then, since $n \in \mathbb{N}$, and any odd number squared minus $1$ is divisible by 4, $a$ will always be a natural number. Is this proof valid, or am I missing something? If it is right, can we define an invertible, bijective mapping $\phi$ such that 
$$\phi : S \to \mathbb{N}$$
Such that $S=\{a_i\}=\{\frac{(2i-1)^2-1}{4}\}_{i=1}^{\infty}$ and $$n=\sqrt{a_i+\sqrt{a_i+\sqrt{a_i+\sqrt{a_i+...}}}}$$ 
 A: I think the main problem with this proof is that it doesn't contain a proof of convergence; actually, it is not clear what $\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}$ is in the first place.
To fix that: for any $a>0$ (integer or not), you may want to define $a_k=\sqrt{a+\sqrt{a+\sqrt{a+\cdots \sqrt{a}}}}$ (with $k$ square roots) - and then define $\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}=\lim_{k\to\infty}a_k$ - if the limit exists!
Now, let's prove the existence of the limit. We have $a_1=\sqrt{a}, a_k=\sqrt{a+a_{k-1}}$. By using induction on $k$, we can easily prove that the sequence is (strictly) monotonously increasing:


*

*Base: $a_2=\sqrt{a+\sqrt{a}}\gt\sqrt{a}=a_1$

*Inductive step: $a_{k+1}=\sqrt{a+a_k}\gt\sqrt{a+a_{k-1}}=a_k$, assuming $a_k\gt a_{k-1}$.


Also, the sequence is bounded from above by $1+2\sqrt{a}$, again by induction:


*

*Base: $a_1=\sqrt{a}\lt 2\sqrt{a}+1$

*Inductive step: $a_{k+1}=\sqrt{a+a_k}\lt \sqrt{a+2\sqrt{a}+1}=\sqrt{a}+1\lt 2\sqrt{a}+1$. 


Thus, by monotone convergence theorem, there is a limit $n=\lim_{k\to\infty}a_k$.

Now we can use the logic you already developed. Because $x\to\sqrt{x}$ is a continuous function:
$$n=\lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{a+a_{k-1}}=\sqrt{a+\lim_{k\to\infty}a_{k-1}}=\sqrt{a+n}$$
so $n=\sqrt{a+n}$ which implies
$$a=n^2-n$$
which yields the solutions for $n$:
$$n=\frac{1\pm\sqrt{1+4a}}{2}$$
Obviously, the solution with the "plus" sign is the correct one. The one with "minus" is smaller than $0$ as we originally assumed $a>0$, so it cannot be the limit of $a_k$.
The fact that $a>0$ implies that $n>1$. Conversely, if $n>1$ then $a=n^2-n=n(n-1)>0$, so the maps $a\to(1+\sqrt{4a+1})/2$ and $n\to n^2-n$ are inverse to each other, as maps from $(0,\infty)$ to $(1,\infty)$ (and vice versa). Those maps let you calculate $n>1$ if you have $a>0$ and calculate $a>0$ if you have $n>1$.

The last ("integer") bit is now simple. If you want integer $n$ and $a$, then first set an integer $n>1$ and then pick $a=n^2-n$, which will automatically be an integer as well.
