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I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone help me see if $\;9\!\cdot\!10^n+4\,$ can be a perfect square ?

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Jan 21, 2023 at 12:34
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    $\begingroup$ I suggest: if $n=2k+1$ it looks like the square root is very close, but not equal to, to $9.5\times 10^k$. Start there. (Note: the problem is easy is $n$ is even. Why?) $\endgroup$
    – lulu
    Jan 21, 2023 at 12:50
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    $\begingroup$ @JustWandering Well, that's a standard Pell's equation. $(m,n)=(6,19)$ is a solution, for instance (there are infinitely many). Can you use that to address the OP's question? $\endgroup$
    – lulu
    Jan 21, 2023 at 13:25
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    $\begingroup$ @lulu, how have you found that solution ? Actually, there does not exist any solution. $\endgroup$
    – Angelo
    Jan 21, 2023 at 14:17
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    $\begingroup$ Dupe of Prove nonsquare $111...161 = (10^k-1)/9+50$. You chose the wrong modulus to test. See this Hint in the dupe for how to choose it. $\endgroup$ Jan 22, 2023 at 0:01

4 Answers 4

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If you reduce mod $11$ you get $(-2)(-1)^n+4 \equiv 2$ or $6 \pmod{11}$. Neither $2$ nor $6$ is a quadratic residue mod $11$.

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    $\begingroup$ Wow, a nice solution ! (+1) $\endgroup$
    – Peter
    Jan 21, 2023 at 13:36
  • $\begingroup$ Congratulations ! I like your solution very much. $\endgroup$
    – Angelo
    Jan 21, 2023 at 14:15
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Note that if $$9\!\cdot\!10^n+4=m^2\implies (m+2)(m-2)=9\!\cdot\!10^n$$

Note that $5^n$ must divide either $m+2$ or $m-2$. If that happens, the rest of the factors are not big enough to maintain the difference of $4$ as $\left|5^n-9\!\cdot\!2^n\right|>4$ for $n\geqslant3$.

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Assume that $9\cdot10^n+4\equiv4$ is a perfect square.
$9\cdot10^n+4\equiv4\pmod9$, so $9\cdot10^n+4$ can be represented as $(9m-2)^2=81m^2-18m+4$ or $(9m+2)^2=81m^2+18m+4$, where $m\in\mathbb{N}$.
If $9\cdot10^n+4=81m^2-18m+4$, $$10^n=9m^2-2m=m(9m-2)$$ This means that $m$ and $9m-2$ must be powers of $10$.
Clearly, $9m-2>m$ because $m>0$. If $m≠1$, $m\equiv0\pmod{10}$, but then $9m-2\equiv8\pmod{10}$, which doesn't work. If $m=1$, $9m-2=7$, which also doesn't work.
If $9\cdot10^n+4=81m^2+18m+4$, $$10^n=9m^2+2m=m(9m-2)$$ This means that $m$ and $9m+2$ must be powers of $10$.
Clearly, $9m+2>m$ because $m>0$. If $m≠1$, $m\equiv0\pmod{10}$, but then $9m+2\equiv2\pmod{10}$, which doesn't work. If $m=1$, $9m+2=11$, which also doesn't work.
Therefore, there is a contradiction and so $9\cdot10^n+4$ cannot be a perfect square.

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Comment:

Lets try construction such a number. Suppose we have:

$k^2-9\times 10^n=4$

We use following known Pell's equation:

$x^2-Dy^1=1$

For $D=10$ we have $x=19$ and $y=6$ such that:

$19^2-10\times 6^2=1$

multiplying both sides by $2^2$ we get"

$38^2-10\times 12^2=4$

we rewrite this as:

$38^2-9\times(4^2\times 10)=4$

multiplying both sides by $25^2$ we get:

$(25\times 38)^2-9\times 10^5=4\times 25^2=2500=2496+4$

Or:

$[(25\times 38)^2-2496]-9\times 10^5=4$

Or generally:

$A=[(25\times 38)\times 10^m-(25\times 10^{2m}+96)]$

here in equation $k^2-9\times 10^n=4$, $n=2m+1$

$A$ must be perfect square.May be by brute force we can find such a number.

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