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Rev

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Smallest number such that concatenation is a square

Challenge

Write a program or function that takes a number \$n\$ and returns the smallest \$k\$ such that concatenation \$n'k\$ is a square. This sequence is described by A071176 on the OEIS.

I/O Examples

input --> output

1   -->    6 (4^2)
10  -->    0 (10^2)
35  -->  344 (188^2)
164 -->  836 (406^2)
284 --> 2596 (1686^2)

Rules

Input will always be a positive decimal integer
Output will be a positive decimal integer with no leading 0's or 0
Standard I/O rules apply
No standard loopholes
This is code-golf, so shortest code in bytes wins

Answer*

# Java, <s>75</s> <s>63</s> 61 bytes

Two bytes less thanks to [Kevin Cruijssen][1].

And another two less in the second version again thanks to [Kevin Cruijssen][1]!

```java
n->{int j=0;for(;Math.sqrt(new Long(n+j))%1>0;j++);return j;}
```
[Try it online!][2]

I also have another version with my first (and very different) approach, which is <s>107</s> <s>92</s> <s>93</s> 91 bytes long:

```java
n->{String a;for(long i=4;!(a=i*i+++"").matches(n+"([1-9].*|0)"););return a.split(n,2)[1];}
```
[Try it online!][3]

Both version have been tested against the first [10000 numbers][4].

  [1]: https://codegolf.stackexchange.com/users/52210/kevin-cruijssen
  [2]: https://tio.run/##LY7BisIwFEXX7Ve8jZDQMVQY3EQLboSBmZVLcRFrWpOpLzF5cRDpt2cqurtwLpxj1U3NnddoT7@5HVSMsHmURSRFpgU7UZHIDKJL2JJxKLbvsdpRMNh/wBeS7nVooIN1xnnzMEhg17XsXGDyR9FZxGsghvoPvh32DCvL@WzR1NJWFZdBUwoIVo5Zlj4dh8n71t@cOcFFGWQv2f4AKvSRT4HF7h5JX4RLJPzEiHVCeT/c2fOxrw@cy7IYyzHnxfLzHw
  [3]: https://tio.run/##LY5BasMwFETX9ikUr6Q4FnEJhaI60E0vkKXx4texEzmyJKSvQEh9dlXFWc0MMzBvgjtUxg56Ot9ir8B78vXMM4@AsidTanlAqfgYdI/SaP79Mp8ndFJfdmTVIxlJE3V1fK6ZgBiNo8okK5uD2FBo5FaWZVkUjM@A/XXwVJcFbevqo@Pb3z0rmGDCDRicJsC9VRKp3r2xtu7EEkVuw49KUC@2u5FnMoPUdH1sOwLu4lmiz04Pj8PMTUBuU4d05GCtetD/RbvvGBN5tuRLjPX74Q8
  [4]: https://oeis.org/A071176/b071176.txt

If this is an answer to a challenge…

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…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one. Explanations of your answer make it more interesting to read and are very much encouraged.
…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

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Edit Summary*

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\$\begingroup\$ You can remove the parenthesis surrounding Math.sqrt(new Long(n+j))%1 to save 2 bytes. \$\endgroup\$
– Kevin Cruijssen
Commented Jan 29, 2020 at 12:53
\$\begingroup\$ @KevinCruijssen True! In an initial version I needed them, but while refactoring I didn't realise I could remove them. Thanks! \$\endgroup\$
– GuestUser
Commented Jan 29, 2020 at 12:59
\$\begingroup\$ Np. Btw, in your second version you can also save 2 bytes: 1 by changing long i=4;String a;while(...); to String a;for(long i=4;...;); to save on the semi-colon, and 1 by removing the + in your regex, since the .* already covers this. Nice answer, btw! And welcome to CGCC! :) \$\endgroup\$
– Kevin Cruijssen
Commented Jan 29, 2020 at 13:07
1

\$\begingroup\$ @KevinCruijssen Thanks again! I totally missed the long inside the for hack. And again the "+" was becuase of a previous version in which I didn't have the ".*|0" part, but whitout that part I didn't cover for the f(10) -> 0 case correctly and other solution with a 0 in them , but when I added it I didn't realise the "+" wasn't necessary any more. Really nice catch :). \$\endgroup\$
– GuestUser
Commented Jan 29, 2020 at 14:16

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