Let $a$ be a perfect square number whose decimal representation is the concatenation of two perfect squares, for example $49$ (from $4$ and $9$), $169$ (from $16$ and $9$), ... and $4900$, $490000$ that are divisible by $10$ (simple). (What is called in mathematics such numbers?)

How we can prove there are infinitely many such perfect squares with above property that not divisible by $10$.

  • $\begingroup$ I might be being dumb but I am really struggling to understand what the property you want is. $\endgroup$ – Jack Yoon Sep 18 '14 at 8:46
  • $\begingroup$ I am not sure what you are exactly asking. Does the sequence $p^2$ with prime $p$ meet your requirement? $\endgroup$ – gammatester Sep 18 '14 at 8:46
  • $\begingroup$ Whatever your question exactly is, the answer is probably yes and you can presumably construct such a sequence using Euclid's formula for Pythagorean triples. $\endgroup$ – J.R. Sep 18 '14 at 8:50
  • $\begingroup$ As I think @MathMan want to ask about those perfect squares which can be obtained after juxtaposition of some perfect squares. e.g. $49$ is obtained after juxtaposition of $4$ and $9$. $\endgroup$ – Shodharthi Sep 18 '14 at 9:01
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    $\begingroup$ See the Online Encyclopedia of Integer Sequences, which lists this as A039686 though it gives remarkably little information oeis.org/A039686 $\endgroup$ – Mark Bennet Sep 18 '14 at 9:48

Let $$x_k = 2^{2k-1} 5^{2k} - 10^k + 1 = 4\underbrace{99...9}_{k-1}\underbrace{00...0}_{k-1}1.$$ Then $x_k^2$ has the desired property, as it is the concatenation of the squares of $$2^{k-1}5^k-1=4\underbrace{99...9}_{k-1} \quad \mbox{ and }\quad 10^k-1 = \underbrace{99...9}_{k}.$$

For small $k$, we get the following:

$$ k=1: \quad x_1^2 = 41^2 = \underbrace{16}_{4^2}\,\underbrace{81}_{9^2}. $$ $$ k=2: \quad x_2^2 = 4901^2 = \underbrace{2401}_{49^2}\,\underbrace{9801}_{99^2}. $$ $$ k=3: \quad x_3^2 = 499001^2 = \underbrace{249001}_{499^2}\,\underbrace{998001}_{999^2}. $$ $$ \vdots $$

It is an easy computation to check that this works for every $k$. Note that this method does not find all solutions, but it does show that there are infinitely many of them. Probably there will be other infinite families of solutions.


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