# Numbers nn such that the digit sum of n2n^2 is a square

Let $$S(n)S(n)$$ be the digit sum of $$n∈Nn\in\mathbb N$$ in the decimal system. About a month ago, a friend of mine taught me the following:

$$S(92)=S(81)=8+1=32S\left(9\color{red}{^2}\right)=S(81)=8+1=3\color{red}{^2}$$
$$S(102)=S(100)=1+0+0=12S\left(10\color{red}{^2}\right)=S(100)=1+0+0=1\color{red}{^2}$$
$$S(112)=S(121)=1+2+1=22S\left(11\color{red}{^2}\right)=S(121)=1+2+1=2\color{red}{^2}$$
$$S(122)=S(144)=1+4+4=32S\left(12\color{red}{^2}\right)=S(144)=1+4+4=3\color{red}{^2}$$
$$S(132)=S(169)=1+6+9=42S\left(13\color{red}{^2}\right)=S(169)=1+6+9=4\color{red}{^2}$$
$$S(142)=S(196)=1+9+6=42S\left(14\color{red}{^2}\right)=S(196)=1+9+6=4\color{red}{^2}$$
$$S(152)=S(225)=2+2+5=32S\left(15\color{red}{^2}\right)=S(225)=2+2+5=3\color{red}{^2}$$

Then, I’ve got the following:

For every $$m∈Nm\in\mathbb N$$, each of the following $$77$$ numbers is a square.
$$S((10(3m−2)2−1)2),S((10(3m−2)2)2),⋯,S((10(3m−2)2+5)2)S\left(\left(10^{(3m-2)^2}-1\right)^2\right),S\left(\left(10^{(3m-2)^2}\right)^2\right),\cdots,S\left(\left(10^{(3m-2)^2}+5\right)^2\right)$$

However, I’m facing difficulty in finding such $$88$$ consecutive numbers. So, here is my question:

Question : What is the max of $$k∈Nk\in\mathbb N$$ such that there exists at least one $$nn$$ which satisfies the following condition?

Condition : Each of the following $$kk$$ numbers is a square.
$$S((n+1)2),S((n+2)2),⋯,S((n+k−1)2),S((n+k)2)S\left((n+1)^2\right),S\left((n+2)^2\right),\cdots,S\left((n+k-1)^2\right),S\left((n+k)^2\right)$$

Note that we have $$k≥7k\ge 7$$. Can anyone help?

Added : A user Peter found the following example of $$k=8k=8$$ :
$$S(460458462)=82,S(460458472)=72,⋯,S(460458522)=72,S(460458532)=82S\left(46045846^2\right)=8^2,S\left(46045847^2\right)=7^2,\cdots,S\left(46045852^2\right)=7^2,S\left(46045853^2\right)=8^2$$
Hence, we have $$k≥8k\ge 8$$.

## Answer

As Djalal Ounadjela outlined in the comments, there is probably no such maximal k, we can expect to find an n for any k at an order of magnitude n ~ 10^m with approximately $m/\log_{10}(m) \sim k$.

PS: http://oeis.org/A061910 lists numbers n for which the sum of digits of n^2 is a square.

Attribution
Source : Link , Question Author : mathlove , Answer Author : Max