On combining $n$ and $n^2$ into one number Consider the sequence $T_n$ formed by combining $n$ and $n^2$ into one number.
ie. (A053061)
$$T_n=\{11,24,39,416,525,636,749 \cdots\}$$
It is easy to see
$$T_n= 10^{\lceil 2 \log_{10}(n) \rceil } n+ n^2$$
I looked at the sequence closely trying to find if there are any perfect squares in the sequence but wasn't able to  upto $n=100$. 
I also was able to prove that if $n^{th}$ term is a perfect square then :
1) $n \equiv 8 \text{ or }  0 (\text{ mod } 9)$
2)In the case where $n=9m$ , m is not a square free number.
But I am unable to attack the question

Do there exist  any perfect square in the sequence?


There are no squares in this sequence.
Elements of this sequence are of the form $b^2+10^n b$ for $10^{n-1} \leq b^2 < 10^n$. So we are looking for a special solution to the equation $a^2 =b^2+ 10^n b$. This is an easy Diophantine equation to solve: We have $a^2 - (b+10^n/2)^2 = - 10^{2n}/4$, or
$(a+b+10^n/2)(a-b-10^n/2) = -10^{2n}/4$
The two factors on the left side have the same parity. Clearly we may assume $n>1$, so these factors both must be even. If those factors are both multiples of $20$, then $a$ and $b$ are both multiples of $10$, and we may divide $a$ and $b$ by $10$ and subtract $1$ from $n$ without changing the equation. However, it does modify the inequality. So each factor is either a power of $2$ or $2$ times a power of $5$, hence the two factors must be $2 \cdot 5^{2n}$ and $2^{2n-3}$. This gives:
$a+b+10^n/2 = 2\cdot 5^{2n}$
$a-b-10^n/2 = 2^{2n-3}$
$b= 5^{2n} - 2^{2n-4} -10^n/2$
For $n=2$, we get $575$, but $575^2>100$. For higher $n$, it's even worse: $b^2$ is much greater than $10^n$. Similarly, multiplying $a$ and $b$ by $10$ and adding $1$ to $n$ also makes $b^2$ larger, relative to $10^n$. So there are no solutions.
