Are there any repeats of a number (eg repeat of 86 is 8686) that are perfect squares?
When we repeat a number, i.e. write it twice in a row, we get another number with even no. of digits. For e.g. repeating '86' gives us the number '8686' Is there any such repeat number which is a perfect square. How many such perfect squares are there?
$\endgroup$ 13 Answers
$\begingroup$Note that repeats of a number are of the form $(10^n+1)x$, where $x$ is the integer that is repeated and $n$ is the number of digits of $x$. Hence repeated numbers are divisible by $10^n +1$ where $n$ is the number of digits of the number that is repeated.
To have a repeated number which is also a square we need to find $n$ such that the product of primes with odd powers in the prime decomposition of $10^n +1$ has value less than $10^n$. Intuitively, the number which is repeated must 'make up' for the odd power primes in the decomposition of $10^n+1.$
This observation proves that there are no perfect square repeated numbers with $4$ digits as $101$ does not satisfy the above requirements ($101$ is prime!). Likewise there are no $6$ digit perfect square repeated numbers as $1001=7\cdot11\cdot13$. To generalise this method for arbitrary $n$ we need knowledge on the prime decomposition of $10^n+1$ in general. I'm not sure if anything is known about this!
Hope this helps!
Edit: Searching online I've found that $$10^{11}+1 = 11^2 \cdot 23 \cdot 4093 \cdot 8779.$$ To form a repeated perfect square, we need $k$ such that $k^2 \cdot 23 \cdot 4093 \cdot 8779 = k^2\cdot826446281$ has $11$ digits. $k=10$ works.
Hence $(10^{11}+1)\cdot100 \cdot 23 \cdot 4093 \cdot 8779=8264462810082644628100$ is a repeated number and a perfect square!
Checking this on wolfram alpha shows that this number is indeed a square. It is the square of $$90909090910.$$
Note that our choice of $k$ was arbitrary. If we instead chose $k = 9$, we arrive at the square repeated number $$6694214876166942148761 = 81818181819^2.$$
$\endgroup$ $\begingroup$There’s a nice trick regarding the decomposition of $10^n+1$ when we want to find a square prime factor.
First,$$1-z^n = (1-z)(z^{n-1}+z^{n-2}+\dots+z^0).$$Set $z=-10$ and $n=2k+1$ giving,$$1-(-10)^{2k+1} = \left(1-(-10)\right)\left((-10)^{2k}+(-10)^{2k-1}+\dots+(-10)^0\right).$$Simplify using even and odds powers. So,$$10^{2k+1} + 1 = 11(10^{2k}-10^{2k-1}+\dots+10^0).$$Thus $11$ is a divisor of $10^{2k+1}+1$. Now look at the remainders of the second factor when divding by $11$\begin{align}\left(10^{2k}-10^{2k-1}+\dots+10^0\right)\bmod{11} &= (-1)^{2k}-(-1)^{2k-1}+\dots+(-1)^0 \\ &= 1 + 1 + \dots + 1 \\ &= 2k+1.\end{align}So when $11$ divides $2k+1$, $11^2$ divides $10^{2k+1} + 1$ and we have our square factor! Note that these are not the only $10^n+1$ with square factors, but the ones with $11^2$ are easy to find.
$\endgroup$ $\begingroup$Here is a sketch of how to find such numbers or prove that they do not exist. I've left some details to be filled in.
Repeating a two digit number, as you have done, leaves a number divisible by $101$ which is a three digit prime, and can't occur as a squared factor of the result (because the square is larger than the number you are interested in).
Duplicating an $n$ digit number leaves something which is divisible by $10^n+1$, and the same observation applies so long as there are no squared factors of $10^n+1$. On the other hand if $10^n+1=p^2q$ then $(10^n+1)q=p^2q^2$
I've deleted what was here because it was wrong. Instead modify the above to choose $t$ so that $r=t^2q$ has $n$ digits. Then $(10^n+1)r=p^2t^2q^2=(ptq)^2$.
Again, you should be able to find a solution with $p=7$ (the lowest available prime), but this will be larger than the solution Zestylemonzi has with $p=11$.
$\endgroup$
