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Here is a math question that I made up, but I don't know how easily it is solved:

Find the smallest positive integer greater than $100$ (or prove that no such number exists) whose digits when read left to right in bases $4,5,$ and $6$ form a perfect square.

Clarification: Since there seem to be many interpretations of what my question was, here is clarification: My original question's intention was to find a base $10$ number that when converted to bases $4,5,$ and $6$ had the digits read from left to right in each of them be a perfect square.

For example, take the number $200_{10} = 3020_{4} = 1300_5 = 532_6$. This doesn't work since not all of $3020,1300,$ and $532$ are perfect squares.

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    $\begingroup$ I assume you rather mean: You look for a digit digit sequence of at least three digits $\in\{0,1,2,3\}$ such that the interpretation of this digit sequence as a base 4 number is a square, the interpretation as a base 5 number is a square, and the interpretation as a base 6 number is a square? (For the fact that an integer is a square would not depend on the base used for its representation; even a roman number such as CXLIV is a square ...) $\endgroup$
    – Hagen von Eitzen
    Commented Dec 9, 2015 at 19:50
  • $\begingroup$ $121$ is such a number, if I have understood the question correctly. $\endgroup$
    – TonyK
    Commented Dec 9, 2015 at 19:51
  • $\begingroup$ Or this: If the solution were $N=42=42_{10}=222_4=132_5=110_6$ then the integers (all in base $10$!) $222$, $132$, $110$ shall be squares? $\endgroup$
    – Hagen von Eitzen
    Commented Dec 9, 2015 at 19:54
  • $\begingroup$ Yes, $25_6$ would be considered a square. $\endgroup$
    – user19405892
    Commented Dec 9, 2015 at 19:56
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    $\begingroup$ I guess there is still a clarification needed: "the digits read from left to right in each of them interpreted in base $10$ be a perfect square." $\endgroup$
    – Hagen von Eitzen
    Commented Dec 9, 2015 at 20:25

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The simplest, if I read the question properly, is $121$. $121_4=25_{10}, 121_5=36_{10}, 121_6=49_{10}$ This reflects $121_b=b^2+2b+1=(b+1)^2$ The same thing works for any polynomial square as long as the base is high enough to avoid overflow, so $144$ is a square in any base that allows $4$, $961$ is a square in any base that allows $9$, and so on.

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  • $\begingroup$ What if I changed the question to greater than $1000$? What would it be then? $\endgroup$
    – user19405892
    Commented Dec 9, 2015 at 19:56
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    $\begingroup$ @user19405892 After your clarification this isn't an answer to your question. And for your question, after $1$ there seems to be no such number at least up to $1000000000$ (or my PC tells me so). $\endgroup$
    – Abstraction
    Commented Dec 9, 2015 at 19:59
  • $\begingroup$ Yes. So I guess this is an open problem now then? Does there exist such a number? $\endgroup$
    – user19405892
    Commented Dec 9, 2015 at 20:02
  • $\begingroup$ You can just add zeros to make it as large as you want. $1000002000001=1000001^2$ in any base $3$ or larger, for example This reflects that $(b^6+1)^2=b^{12}+2b^6+1$ $\endgroup$
    – Ross Millikan
    Commented Dec 9, 2015 at 20:03
  • $\begingroup$ My original question's intention was to find a base 10 number that when converted to base $4,5,$ and $6$ had the digits read from left to right in each of them be a perfect square. $\endgroup$
    – user19405892
    Commented Dec 9, 2015 at 20:08

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