# When is the concatenation of numbers a square?

A while back I asked a question concerning the existence of squares of the form $1234567891011\ldots$ This question is similar but more general. Let $n,p,\ell$ be positive integers, then we define $c(n,p,\ell)$ to be the concatenation of numbers starting with $n$, having a period $p$, and length (or number of steps) $\ell$, then we might ask ourselves - When is $c(n,p,\ell)$ square? For example $$c(1,p,0)=1,$$ $$c(4,p,0)=4,$$ $$c(1,20,1)=121,$$ $$c(10,14,1)=1024.$$ A non-example: $$c(1,2,5)=1357911.$$ So, it's easy to find numbers with $\ell\leqslant 1$ being square, but I have not been able to find any square of the form $c(n,p,2)$. Is there such a number? In particular I'd like to know whether there is any literature relating to this, or some similar problem.

UPDATE: Gerry Myerson found a number satisfying the requirements! For completeness I will list the numbers $c(n,p,2)$ with $n<20$, $p<20000$ below.

$c(1,224,2)=1225449=1107^2$
$c(2,7,2)=2916=54^2$
$c(5,28,2)=53361=231^2$
$c(6,465,2)=6471936=2544^2$
$c(13,13694,2)=131370727401=362451^2$
$c(14,2371,2)=1423854756=37734^2$
$c(16,4154,2)=1641708324=40518^2$

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what do you mean by steps (l)? can you give an example with c(,,2)? – j13r Apr 21 '12 at 10:17
@j13r: I have edited the post. – Carolus Apr 21 '12 at 10:21

## 1 Answer

$c(5,28,2)=53361=(231)^2{}{}$.

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