81 is a square number. Find the minimum number of zeros needed to be inserted between 8 and 1 such that it makes another perfect square, or prove that it is impossible.

Suppose there does exist some \( n \) where \( n^2 = 800...001 \). Then since \( 800..001 \) is odd, \( n \) must. Therefore, \( n = 2k + 1 \). Then,

\( (2k + 1)^2 = 800...001 \)

\( 4k^2 + 4k + 1 = 800..001 \)

\( 4k(k+1) = 800...001 \)

\( k(k+1) = 200...000 \)

\( k(k+1) = 2 * 2^t*5^t \) for some \( t > 1 \)

\( k(k+1) = 2^{t+1}*5^t \)

Now, \( \gcd(k,k+1) = 1 \).Also, since \( 5^t > 2^{t+1} \), we must have \( k+1 = 5^t \) and \( k = 2^{t+1} \).
But, we can easily prove that \( 5^t > 2^{t+1} + 1 \) for \( t > 1 \). Therefore, we have a contradiction, and there exist no integer \( n \) for which \( n^2 = 800...001 \)

@Bryan Lee Shi Yang
–
Sorry , but did you mean that you salute me ? It's actually Siddhartha who posted the solution , so I guess you should be saluting him :)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestNot possible.

Suppose there does exist some \( n \) where \( n^2 = 800...001 \). Then since \( 800..001 \) is odd, \( n \) must. Therefore, \( n = 2k + 1 \). Then,

\( (2k + 1)^2 = 800...001 \)

\( 4k^2 + 4k + 1 = 800..001 \)

\( 4k(k+1) = 800...001 \)

\( k(k+1) = 200...000 \)

\( k(k+1) = 2 * 2^t*5^t \) for some \( t > 1 \)

\( k(k+1) = 2^{t+1}*5^t \)

Now, \( \gcd(k,k+1) = 1 \).Also, since \( 5^t > 2^{t+1} \), we must have \( k+1 = 5^t \) and \( k = 2^{t+1} \). But, we can easily prove that \( 5^t > 2^{t+1} + 1 \) for \( t > 1 \). Therefore, we have a contradiction, and there exist no integer \( n \) for which \( n^2 = 800...001 \)

Log in to reply

Nice solution :)

Log in to reply

Nice solution

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

Log in to reply