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Clearly, $3^2=9$ is a perfect square, all of whose digits are $7$, $8$, or $9$. Are there any other perfect squares with this property?

This is an interesting question that does not seem to be solved yet, coming from AoPS (https://artofproblemsolving.com/community/c6h1928519). As duck_master seems to show, it should be impossible to solve this problem by analyzing the quadratic residues modulo $10^n$ for some $n$.

I strongly suspect the answer is no. I have been running a Python script for quite some time, and it has checked squares up to $(50,000,000,000)^2$ with no results (unless I messed up the code).

greenturtle3141
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  • As a minimum it has to be a number with the lowest order digit 3 or 7. These squares have the lowest digit 9. There is no way to end in 7 or 8. – herb steinberg Oct 16 '19 at 03:56
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    The number to be squared has to be $\equiv\pm17\pmod{50}$ for the two last digits to be a legal combo. I'm sure the folks at AOPS have checked it higher. I suspect $p$-adic techniques to prove that congruences modulo $10^n$ won't settle this. – Jyrki Lahtonen Oct 16 '19 at 04:05
  • I would suggest you also add in the example which you've given in the AoPS forum. I had to go there to get a little grasp of what the problem was. – user712576 Oct 16 '19 at 15:32
  • Is the choice an arbitrary one? What about $5,6,7$. Can they exhibit the same property as well? – user712576 Oct 16 '19 at 16:30
  • @fishfag at least for $5,6,7$ we have $26^2=676$. Perhaps the motivation for the choice of $7,8,9$ is that $6,7,8,9$ does not work. Though, as Jyrki's answer suggests, the case $8,9$ may not be so simple either. – greenturtle3141 Oct 16 '19 at 17:22
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    And then there are also numbers like $7917^2=62678889$, which really close in but fails at the first digits. Another way of looking at the problem is that a positive integer made up of only $7,8 or 9$ cannot be expressed as the sum of consecutive odd numbers. – user712576 Oct 16 '19 at 17:32
  • Consider the following crude probabilistic model: For $m$ in the range $[10^{(n-1)/2},10^{n/2}]$ its square has $n$ digits, hence $m^2$ qualifies with probability $(3/10)^n$. Thus the expected number of positive answers is $$\sum_{n=1}^\infty(\frac3{10})^n(10^{n/2}-10^{(n-1)/2})\approx 12.64$$ Either the truth is unlucky, or my model is off in some crucial way :-) – Jyrki Lahtonen Oct 18 '19 at 20:04
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    Interesting near-miss: 8943158167 ^ 2 = 79980077999978799889 – Brian Trial Apr 16 '21 at 20:06

3 Answers3

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For $n<10^{26}$, It seems only $$9949370777987917^2=98989978877879888789778997998889$$ with digit 789.

Semiclassical
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Northwolves
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  • This is also the smallest solution posted here (includes many other 3-digit patterns, too): http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/math0210.htm#789. – MaoWao Feb 26 '24 at 13:19
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A short proof for the fact that $\ldots88889$ will appear as the last decimal digits of a square. Consider the modular inverse of $3$ modulo $10^m$. That is, let $n\equiv\ldots 66667$. Then $(3n)^2\equiv1\pmod {10^m}$ and therefore $n^2$ is the modular inverse of $9$. Modulo $10^m$ we have $-1/9\equiv\ldots11111$, so we also have $$n^2\equiv(1/3)^2=1/9\equiv-(\ldots11111)\equiv\ldots88889\pmod{10^m}.$$

Of course, this does not settle the main question, only proving the futility of trying to prove the non-existence of such squares by studying any finite segment of least significant digits.

Jyrki Lahtonen
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  • In other words, check out the last digits of $7^2$, $67^2$, $667^2$, $6667^2$,... – Jyrki Lahtonen Oct 16 '19 at 04:16
  • You've proved that $\ldots66667$, when squared, ends in $\ldots88889$. Can you clarify how this may help in the big picture? Are you claiming that if a perfect square has only the digits $7$, $8$, and $9$, then it must necessarily end in $\ldots88889$? – greenturtle3141 Oct 16 '19 at 04:46
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    @greenturtle3141 Unfortunately no. The guy on AOPS explained why the three last digits of the square must be $889$. But for example $\ldots7889$ is not ruled out. For example $167^2=27889$. I only wanted to give a "local" argument showing that squares ending with $\ldots88889$ exist for any number of $8$s. This is worthless for the purposes of the main question. – Jyrki Lahtonen Oct 16 '19 at 04:53
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COMMENT: We can work on 'construction' of such a number.I propose following method which is based on optimization of digits; consider following experiment. Let the number of digits which are not 7, 8 or 9 be x. Consider 7 digits number $d_1d_2d_3d_4d_5d_6d_7$; we start with number $3161517$, we have:

$3161517^2=995189741289$; x=5

Where $d_3=6$, we try 7 for this digit:

$3171717^2=10059788728089$; x=6

So $d_3<7$ is more desirable; now we work on $d_4$ and start with :

$3162617^2=10002146288689$; x=7

So $d_4<2$; we try $d_4=0$:

$3160617^2=9989499820689$; x=4

So $d_4=0$ is best. Now we work on $d_5$:

$d_5=7$ gives $3160717^2=9990131954089$; x=7

$d_5=$ gives $3160517^2=9988867707289$; x=3

With $d_5=4$; x=5 and with $d_5=3$; x=7, so optimum digits for minimum x are:

$d_1=3, d_2=1, d_3=6, d_4=0, d_5=5, d_6=1, d_7=7$ and optimized number is $3160517$.

For example based on this method I found $88317^2=7799892489$ with x=2. Surely we can find numbers with x=1. I think finding a number with x=0 is probable.An efficient computer program can surely help.

sirous
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