Find all five digit number $\overline{abcde}$ such that $$\overline{abcde} = \overline{(ace})^2$$

This question popped in my mind while solving other elementary numbers and I have been trying to solve it ever since but without any luck.

My Take : Since the digit place of $e^2$ should be equal to digit's place of $e$ , So the only possible values of $e$ are $0,1,5$ and $6$

Also since the First digit of the numbers are equal , We can conclude that the the only possible values of $a=1$.

Hence our number can take the following form :


But how do we further solve this?

Also Another interesting part of this question would be to solve for $\overline{abcd}$ such that $$\overline{abcd} = \overline{(bd)}^2$$

  • $\begingroup$ Given that $a=1$, at worst you have to search the numbers from $100$ to $199$. Quite a lot fewer as the last digit is also constrained. $\endgroup$ – lulu Oct 5 '19 at 11:13
  • $\begingroup$ You are right , but are there any more logical or mathematical ways to reduce the possibilities further? $\endgroup$ – The Demonix _ Hermit Oct 5 '19 at 11:18
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    $\begingroup$ There only seem to be three cases. For example you can exclude anything above $100\sqrt{2}$ $\endgroup$ – Henry Oct 5 '19 at 11:19
  • $\begingroup$ in fact lulu using PARI/GP you can show there are only 5 numbers in the whole of the three digit range, that have squares that leave 2 other digits to be $b$ and $d$ in any order. $\endgroup$ – user645636 Oct 5 '19 at 11:21
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    $\begingroup$ As others have remarked, the search is really extremely narrow. This sort of problem often comes down to some case work. $\endgroup$ – lulu Oct 5 '19 at 11:23

We can minimize trial and error with some clever use of modular arithmetic.

Let $N=100a+10c+e$ be the square root. Thus $N^2\equiv e^2$ and we require also $N^2\equiv e\bmod 10$. Therefore $e^2\equiv e$ forcing $e\in\{0,1,5,6\}$.

We also know that $(100a)^2<10000(a+1)$ or $a^2<a+1$ forcing $a=1$. Then $N^2<20000$ but $145^2>140×150=21000$, therefore $N<145$. This result together with the earlier constraint on $e$ leaves only eighteen candidates, which can be exhaustively searched with little trouble; but we can do even better than that.

Consider the case $e=0$. Then $N=100+10c$ (with $a=1$) and $N^2=10000+2000c+100c^2$. For the hundreds digit in $N^2$ to match $c$ we must then have $c^2\equiv c\bmod 10$. This constraint admits$c\in\{0,1,5,6\}$, but only $0$ and $1$ satisfy the bounty $N<145$ which implies $c\le 4$. Thereby we identify



For $e=1$ we have


With $c\le 4$, $20c+1<100$ and thus the hundreds digit is $\equiv c^2+2\bmod 10$. Therefore we must satisfy

$c^2-c+2\equiv 0\bmod 10$

which has a discriminant that is not a quadratic residue $\bmod 5$. So nobody's home here.

The cases $e=5$ and $e=6$ are left to the reader; they are handled similarly to $e=1$ as described above. For these cases $N<145$ implies $c\le 3$ which will then fix the hundreds digit as $\equiv c^2+c$ ($e=5$) or $\equiv c^2+c+2$ ($e=6$). We will then get only one additional solution which the reader can find. I list the complete solution set as (with $x$ digits to be filled in):





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