6
$\begingroup$

I have trouble proving if $10^m + 10^n+10^p+1$ can be a square number for $m>n,>p\geq 1$ integers.

I have tried some modulos against $8$ and $9$ and other prime number, because $10^m + 10^n+10^p+1$ can be 1 (mod 8) when $p \geq 3$ and $10^m + 10^n+10^p+1$ always be 4 (mod 9), and $10^m + 10^n+10^p+1$ can be 4 (mod q) for any prime number $q \geq 7$, which can't give me the conclusion $10^m + 10^n+10^p+1$ will be a square number or not.

For example, $1049^2=1100401$, very close to $10^m + 10^n+10^p+1$, so I think it will take more research to prove this problem.

$\endgroup$
1
  • 2
    $\begingroup$ There are many squares of this form, if we allow $n=p$. This admittedly silly observation means that any argument needs to have a step, where $n>p$ is crucial. Modulo any integer we can find occasions when $10^n\equiv 10^p$ even though $n>p$. So congruences alone won't do. $\endgroup$ Commented Jul 10 at 5:27

1 Answer 1

10
$\begingroup$

No, $(10^m + 10^n + 10^p + 1)$ can never be a perfect square.

Here, $p$ has two conditions: either $ p = 1 $ or $ p > 1 $.

  1. For $ p = 1 $: The number ends in 11. For a perfect square ending in 1, the second last digit needs to be even, which contradicts the form $ 10^m + 10^n + 10^p + 1 $.

  2. For $ p > 1 $: The number ends in 01. Only numbers ending in 1 and 9 can be perfect squares with the last digit of 1.

Let's analyze both cases:

  • Case Last Digit $= 1 $:

    $(10x + 1)(10x + 1) = 100x^2 + 20x + 1$

    To fit this into $ 10^m + 10^n + 10^p + 1 $, $ x $ needs to be a multiple of 5 (i.e., $ x = 5a $). However, $x^2 = 25a^2$

    which does not satisfy the form $ 10^m + 10^n + 10^p + 1 $.

  • Case Last Digit $ = 9 $:

    $(10x - 1)(10x - 1) = 100x^2 - 20x + 1 $

    This expression results in a number ending in 1. To be a perfect square and fit the form $ (10^m + 10^n + 10^p + 1 )$, $ (100x^2 - 20x + 1 )$ would need to match the structure where $ m, n, p $ are exponents of 10.

In both scenarios, $ 10^m + 10^n + 10^p + 1 $ cannot form a perfect square because the conditions required for $ p = 1 $ and $ p > 1 $ do not align with the requirements for a perfect square's digit structure. Thus, $ 10^m + 10^n + 10^p + 1 $ remains non-square for any valid integer values of $ m, n, $ and $ p $.

$\endgroup$
9
  • $\begingroup$ Actually, I have a question, why $x^2=25 a^2$, $100 x^2+20 x+1 = 2500 a^2 + 100 a + 1$ can't not satisfy the form $10^m+10^n+10^p+1$ ? For example, when if $a=21$, $2500 a^2 + 100 a + 1 = 1104601$ , and if $a=210$, $2500 a^2 + 100 a + 1 = 110271001$, which can not tell what kind of digits $2500 a^2 + 100 a + 1$ have. $\endgroup$
    – uk702
    Commented Jul 10 at 2:11
  • $\begingroup$ To satisfy $100a$, $a$ needs to be $10^b$. To satisfy $2500a^2$, $a$ needs to be $2 \cdot 10^b$. These requirements contradict each other, as $a$ cannot be both $10^b$ and $2 \cdot 10^b$. $\endgroup$ Commented Jul 10 at 3:13
  • $\begingroup$ An obvious generalization of this is whether there exists any number (besides powers of 100) whose decimal representation only contains 0s and 1s. This has been considered previously at MSE here and MathOverflow here. $\endgroup$ Commented Jul 10 at 4:40
  • $\begingroup$ Let $a = (\overline{e....gf}) * 10^k$, which $2500 a^2 + 100 a + 1 = 25 * (\overline{e...gf})^2 * 10^{2k+2} + (\overline{e...gf}) * 10^{k+2} + 1$ Because $2500 a^2 + 100 a + 1$ must be star with $10$ or $110$, so digit $e$ can be $2$ or $6$. $\endgroup$
    – uk702
    Commented Jul 10 at 6:46
  • 1
    $\begingroup$ Good catch---yes, that was my intention. @J.W.Tanner $\endgroup$ Commented Jul 10 at 22:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .