Fun question:
Prove that a natural number formed by only 6's and 0's can never be a perfect square.

Question

Fun question: 
Prove that a natural number formed by only 6's and 0's can never be a perfect square.

ganeshie8 · Answer

long time! good to see you again :)

shubhamsrg · Answer

Just came back to check if anything has changed. Woah ! the site became awesom-er ! Good to be back. ^_^
You have been a prominent member ever-since it seems :D

ganeshie8 · Answer

counter example 
$$0 = 0^2$$

shubhamsrg · Answer

question edited.

ganeshie8 · Answer

Since $n^2 \equiv 0,1 \pmod{4}$, we may try something like this :

$$\sum\limits_{k} a_k*10^k\equiv   \sum\limits_{k=0}^1 a_i2^k \pmod 4 $$

because the terms are divisible by $4$  for $k\ge 2$

ganeshie8 · Answer

Ahh that is just divisibility by 4 rule, its not going to help much...

shubhamsrg · Answer

This was a fantastic link that you added ! An interesting approach.

ganeshie8 · Answer

$$60060600 \= 6*10010100 \= 6*100101*10^2 \= 3*100101*2^{\color{red}{3}}*5^{\color{red}{2}}$$

I think it is easy to show that the exponents of $2$ and $5$ cannot be both even.

shubhamsrg · Answer

Spot on.

kainui · Answer

Here's my busted up excuse for a proof. It's complete :P

If it's made up of only 6 and 0s then the leading digit must be a 6.

Since it's square, some number's gotta square to it, the only way that's possible is if the leading digit of its square root is 8, since $8^2=64$ is the only option.

To get rid of that 4 we have to add 2 (since although adding 6 would turn this digit to 0, it would turn our 6 above into 7). In other words the first two digits are 6s:

$$66... = (8X...)^2$$

Here's where it breaks! It looks like puke if you write it out 'officially' so instead of wasting time typing up latex, we know X must be some value between 1 and 9, so quick trial and error gives that:

X=1 $\implies 65...$ 
X=2 $\implies 67...$

And of course higher X just give 67 or higher, so we can't get 6 in that digit!

ganeshie8 · Answer

How do we know the leading digit of the root is 8 ?

kainui · Answer

No other numbers square to 6 in the leading digit. We can at most carry a 1 from lower digits, so if there was a digit that squared to a leading digit of 5 we might have had something to work with. But there isn't, so we don't.

kainui · Answer

Actually I think you're right I've overlooked somethin

kainui · Answer

Oh well, this is too trial-and-error based of a method for me to care to try to fix it up, but I think ultimately there might be a few more cases to check since it looks like there can be up to 2 carried over so 7 might be the leading digit. 

At any rate, this sorta idea will work as long as the statement to be proved is true I think, so I'm not too worried about fixing it, I just wanted to have fun coming up with an idea. :P

kainui · Answer

It would be interesting if this sorta algorithm never terminated and there were infinite-digit numbers made up of 6s and 0s that ARE perfect squares though...