Subfield: Algebra


Question: Determine whether there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers which satisfies the equality

$$
a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}}
$$

for every positive integer $n$.

Solution:
The answer is no.

Suppose that there exists a sequence $\left(a_{n}\right)$ of positive integers satisfying the given condition. We will show that this will lead to a contradiction.

For each $n \geq 2$ define $b_{n}=a_{n+1}-a_{n}$. Then, by assumption, for $n \geq 2$ we get $b_{n}=\sqrt{a_{n}+a_{n-1}}$ so that we have

$$
b_{n+1}^{2}-b_{n}^{2}=\left(a_{n+1}+a_{n}\right)-\left(a_{n}+a_{n-1}\right)=\left(a_{n+1}-a_{n}\right)+\left(a_{n}-a_{n-1}\right)=b_{n}+b_{n-1} .
$$

Since each $a_{n}$ is a positive integer we see that $b_{n}$ is positive integer for $n \geq 2$ and the sequence $\left(b_{n}\right)$ is strictly increasing for $n \geq 3$. Thus $b_{n}+b_{n-1}=\left(b_{n+1}-b_{n}\right)\left(b_{n+1}+b_{n}\right) \geq b_{n+1}+b_{n}$, whence $b_{n-1} \geq b_{n+1}$ - a contradiction to increasing of the sequence $\left(b_{i}\right)$.

Thus we conclude that there exists no sequence $\left(a_{n}\right)$ of positive integers satisfying the given condition of the problem.
Suppose that such a sequence exists. We will calculate its members one by one and get a contradiction.

From the equality $a_{3}=a_{2}+\sqrt{a_{2}+a_{1}}$ it follows that $a_{3}>a_{2}$. Denote positive integers $\sqrt{a_{3}+a_{2}}$ by $b$ and $a_{3}$ by $a$, then we have $\sqrt{2 a}>b$. Since $a_{4}=a+b$ and $a_{5}=a+b+\sqrt{2 a+b}$ are positive integers, then $\sqrt{2 a+b}$ is positive integer.

Consider $a_{6}=a+b+\sqrt{2 a+b}+\sqrt{2 a+2 b+\sqrt{2 a+b}}$. Number $c=\sqrt{2 a+2 b+\sqrt{2 a+b}}$ must be positive integer, obviously it is greater than $\sqrt{2 a+b}$. But

$$
(\sqrt{2 a+b}+1)^{2}=2 a+b+2 \sqrt{2 a+b}+1=2 a+2 b+\sqrt{2 a+b}+(\sqrt{2 a+b}-b)+1>c^{2}
$$

So $\sqrt{2 a+b}<c<\sqrt{2 a+b}+1$ which is impossible.
We will show that there is no sequence $\left(a_{n}\right)$ of positive integers which consists of $N>5$ members and satisfies

$$
a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} \tag{1}
$$

for all $n=1, \ldots, N-2$. Moreover, we will describe all such sequences with five members.

Since every $a_{i}$ is a positive integer it follows from (1) that there exists such positive integer $k$ (obviously $k$ depends on $n$ ) that

$$
a_{n+1}+a_{n}=k^{2} \tag{2}
$$

From (1) we have $\left(a_{n+2}-a_{n+1}\right)^{2}=a_{n+1}+a_{n}$, consider this equality as a quadratic equation with respect to $a_{n+1}$ :

$$
a_{n+1}^{2}-\left(2 a_{n+2}+1\right) a_{n+1}+a_{n+2}^{2}-a_{n}=0
$$

Obviously its solutions are $\left(a_{n+1}\right)_{1,2}=\frac{2 a_{n+2}+1 \pm \sqrt{D}}{2}$, where

$$
D=4\left(a_{n}+a_{n+2}\right)+1 \tag{3}
$$

Since $a_{n+2}>a_{n+1}$ we have

$$
a_{n+1}=\frac{2 a_{n+2}+1-\sqrt{D}}{2}
$$



From the last equality, using that $a_{n+1}$ and $a_{n+2}$ are positive integers, we conclude that $D$ is a square of some odd number i.e. $D=(2 m+1)^{2}$ for some positive integer $m \in \mathbb{N}$, substitute this into (3):

$$
a_{n}+a_{n+2}=m(m+1) \text {. } \tag{4}
$$

Now adding $a_{n}$ to both sides of (1) and using (2) and (4) we get $m(m+1)=k^{2}+k$ whence $m=k$. So

$$
\left\{\begin{array}{l}
a_{n}+a_{n+1}=k^{2} \\
a_{n}+a_{n+2}=k^{2}+k
\end{array}\right. \tag{5}
$$

for some positive integer $k$ (recall that $k$ depends on $n$ ).

Write equations (5) for $n=2$ and $n=3$, then for some positive integers $k$ and $\ell$ we get

$$
\left\{\begin{array}{l}
a_{2}+a_{3}=k^{2}, \\
a_{2}+a_{4}=k^{2}+k, \\
a_{3}+a_{4}=\ell^{2}, \\
a_{3}+a_{5}=\ell^{2}+\ell .
\end{array}\right. \tag{6}
$$

Solution of this linear system is

$$
a_{2}=\frac{2 k^{2}-\ell^{2}+k}{2}, \quad a_{3}=\frac{\ell^{2}-k}{2}, \quad a_{4}=\frac{\ell^{2}+k}{2}, \quad a_{5}=\frac{\ell^{2}+2 \ell+k}{2} . \tag{7}
$$

From $a_{2}<a_{4}$ we obtain $k^{2}<\ell^{2}$ hence $k<\ell$.

Consider $a_{6}$ :

$$
a_{6}=a_{5}+\sqrt{a_{5}+a_{4}}=a_{5}+\sqrt{\ell^{2}+\ell+k} .
$$

Since $0<k<l$ we have $\ell^{2}<\ell^{2}+\ell+k<(\ell+1)^{2}$. So $a_{6}$ cannot be integer i.e. there is no such sequence with six or more members.

To find all required sequences with five members we must find positive integers $a_{2}, a_{3}, a_{4}$ and $a_{5}$ which satisfy (7) for some positive integers $k<\ell$. Its clear that $k$ and $\ell$ must be of the same parity. Vise versa, let positive integers $k, \ell$ be of the same parity and satisfy $k<\ell$ then from (7) we get integers $a_{2}, a_{3}, a_{4}$ and $a_{5}$ then $a_{1}=\left(a_{3}-a_{2}\right)^{2}-a_{2}$ and it remains to verify that $a_{1}$ and $a_{2}$ are positive i.e. $2 k^{2}+k>\ell^{2}$ and $2\left(\ell^{2}-k^{2}-k\right)^{2}>2 k^{2}-\ell^{2}+k$.
It is easy to see that $\left(a_{n}\right)$ is increasing for large enough $n$. Hence

$$
a_{n+1}<a_{n}+\sqrt{2 a_{n}} \tag{1}
$$

and

$$
a_{n}<a_{n-1}+\sqrt{2 a_{n-1}} . \tag{2}
$$

Lets define $b_{n}=a_{n}+a_{n-1}$. Using AM-QM inequality we have

$$
\frac{\sqrt{2 a_{n}}+\sqrt{2 a_{n-1}}}{2} \leq \sqrt{\frac{2 a_{n}+2 a_{n-1}}{2}} \tag{3}
$$

Adding (1), (2) and using (3):

$$
b_{n+1}<b_{n}+\sqrt{2 a_{n}}+\sqrt{2 a_{n-1}} \leq b_{n}+2 \sqrt{b_{n}}
$$

Let $b_{n}=m^{2}$. Since $\left(b_{n}\right)$ is increasing for large enough $n$, we have:

$$
m^{2}<b_{n+1}<m^{2}+2 m<(m+1)^{2}
$$

So, $b_{n+1}$ can't be a perfect square, so we get contradiction.