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

S(92)=S(81)=8+1=32

S(102)=S(100)=1+0+0=12

S(112)=S(121)=1+2+1=22

S(122)=S(144)=1+4+4=32

S(132)=S(169)=1+6+9=42

S(142)=S(196)=1+9+6=42

S(152)=S(225)=2+2+5=32Then, I’ve got the following:

For every m∈N, each of the following 7 numbers is a square.

S((10(3m−2)2−1)2),S((10(3m−2)2)2),⋯,S((10(3m−2)2+5)2)However, I’m facing difficulty in finding such 8

consecutivenumbers. So, here is my question:

Question: What is the max of k∈N such that there exists at least one n which satisfies the following condition?

Condition: Each of the following k numbers is a square.

S((n+1)2),S((n+2)2),⋯,S((n+k−1)2),S((n+k)2)Note that we have k≥7. Can anyone help?

Added: A user Peter found the following example of k=8 :

S(460458462)=82,S(460458472)=72,⋯,S(460458522)=72,S(460458532)=82

Hence, we have k≥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/log10(m)∼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*