In this question, the questioner asked to prove that $$f(n)=\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$$ is a perfect square for all $n\in\mathbb N$. However, I was not able to find any $n$ for which this is true. So I have the following "counter-conjecture": For any $n\in\mathbb N=\{1,2,3,4,\dots\}$, $f(n)$ is not a perfect square.

How can I prove this?

Notice that by simply evaluating the geometric series we get $$f(n)=\frac{8}{9} \left(10^n-1\right)+\frac{4}{9} \left(10^{2 n}-1\right)+4=\frac{4}{9} \left(2^{n+1} 5^n+100^n+6\right),$$

so it would be enough to prove that $$\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$$ is never an integer.

But I don't know how to do this.

  • 2
    $\begingroup$ what about $n=0$? $\endgroup$ Dec 10 '19 at 21:24
  • 1
    $\begingroup$ Hi @J.W.Tanner, I use the convention that $0\not\in\mathbb N$. I‘ve also edited the question $\endgroup$
    – user729882
    Dec 10 '19 at 21:27
  • 2
    $\begingroup$ $100^n+2^{n+1}5^n+6=(10^n+1)^2+5$ $\endgroup$
    – saulspatz
    Dec 10 '19 at 21:29
  • 2
    $\begingroup$ @saulspatz then you must to prove $(10^n+1)^2+5 \neq m^2$ $\endgroup$ Dec 10 '19 at 21:32
  • 1
    $\begingroup$ Also, one can note that $$f(n)\equiv 8\pmod{16}$$ which is impossible for square numbers $\endgroup$ Dec 10 '19 at 21:33

For all naturals $n$ we have $$f(n+1)-f(n)=44\cdot10^{2n}+8\cdot{10^n}.$$

Hence, $$\frac{f(n+1)-f(n)}{16}=11\cdot25\cdot10^{2(n-1)}+5\cdot10^{n-1}\in\mathbb N.$$

It follows that $$f(n)\equiv f(1)\equiv8\pmod{16}$$ for all $n\in\mathbb N$.

Because every square is equivalent to $0,1,4$ or $9$ modulo $16$, it follows that $f(n)$ is never a square.



Can you show that $\sqrt{f(n)}$ is between consecutive integers $\dfrac2310^n+\dfrac13$ and $\dfrac2310^n+\dfrac43$?


Whenever I see something of the form $A + 2A\times\text{something of magnitude} + A\times \times\text{something of geometrically more magnitude}$, I figure it fits roughly into the form of $A(\text{something of mangitude} + 1)^2$.

So so $4_{2n}= \frac 49\times (10^{2n} -1)$ is roughly of magnitude $(10^n)^2$ and $8_{n} = 49\times (10^n -1)$ is roughly of magnitude $10^n$ then we should figure $4_{2n}+ 8_n + 4$ is fairly close so $\frac 49(10^n + 1)^2$.

And it turns it is VERY close (but slightly more) that it is too large to be a square of $\frac 23(10^n+1)$ but too small to be the next integer square.

(We might pause and consider that $\frac 23(10^n+1)$ need not be an integer. That turns out not to be relevant.)


To continue along your lines:

Express as $\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$

$=\frac 23\sqrt{(10^n)^2 + 2*10^n + 6}$

Notice that $n \ge 1$ then $(10^n + 1)^2 = 10^{2n} + 2*10^n + 1 < (10^n)^2 + 2*10^n + 6 < 10^{2n} + 4*10^n + 4 = (10^n + 2)^2$

So $(10^n)^2 + 2*10^n + 6$ is not a perfect square (but is an integer) so $\sqrt{2^{n+1} 5^n+100^n+6}$ is irrational and $\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$ is not an integer.


Your Answer

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