10
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – herb steinberg Oct 16 '19 at 3:56
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 16 '19 at 4:05
  • $\begingroup$ 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. $\endgroup$ – user712576 Oct 16 '19 at 15:32
  • $\begingroup$ Is the choice an arbitrary one? What about $5,6,7$. Can they exhibit the same property as well? $\endgroup$ – user712576 Oct 16 '19 at 16:30
  • 1
    $\begingroup$ 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. $\endgroup$ – user712576 Oct 16 '19 at 17:32
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In other words, check out the last digits of $7^2$, $67^2$, $667^2$, $6667^2$,... $\endgroup$ – Jyrki Lahtonen Oct 16 '19 at 4:16
  • $\begingroup$ 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 $\ldots88889$? $\endgroup$ – greenturtle3141 Oct 16 '19 at 4:46
  • 2
    $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Oct 16 '19 at 4:53
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.