Another BIG number

[sagan_fan]

I have a feeling that the number 19683^(387420489)-2 is either a square or on the form a^a. Anybody wanna help me prove it?

ps. If anybody doesnt like me asking non-mersenne questions, then just say so. I picked this forum because it seems to be the best around, maybe it was wrong.

[svempasnake]

Originally Posted by sagan_fan

I have a feeling that the number 19683^(387420489)-2 is either a square or on the form a^a. Anybody wanna help me prove it?

Hi, sounds like a fun challange helping you finding that out! But I don't know yet how to do that in practice. :( However, I think I can tell that your number is about 1,663,618,948 decimal digits, or about 690,802,815 * 8 bits. I.e. almost within reach for arithmetic operations in RAM on a modern PC! Btw, I think your number the same as (3^9) ^ (3^18 ) - 2 ?

[sagan_fan]

Yes, it is, that was the way i found the number, I just thought it might be better to write in exponential form.

[cperciva]

Originally Posted by sagan_fan

I have a feeling that the number 19683^(387420489)-2 is either a square or on the form a^a. Anybody wanna help me prove it?

No, but I'll disprove it for you:

200390687*ln(200390687.) = .38306241773354074609 * 10^10
387420489*ln(19683.) = .38306241908748739381 * 10^10
200390688*ln(200390688.) = .38306241974511869174 * 10^10

Since x*ln(x) is monotonically increasing, this proves that your number is not in the form a^a.

Further,

19683 ^ 387420489 - 2 mod 29 = 19

and the quadratic residues modulo 29 are 0,1,4,5,6,7,9,13,16,20,22,23,24,25,28; so your number is not a square either.

It is possible that it could be some other prime power; but if it were, it would grossly violate the ABC conjecture.

[sagan_fan]

I don't really understand your proof. Where did 200390688 come from? What's an ABC conjecture?
Anyway, i guess I have to trust you. Looks like i've gotta rewrite my theories a little. ops: Back to square one.
Thanks for the help.

[binarydigits]

Originally Posted by sagan_fan

Where did 200390688 come from?

cperciva is saying that your number is somewhere between 200390687^200390687 and 200390688^200390688 so therefore cannot be of the form a^a.

[sagan_fan]

Oh, yeah,now I see. He just took the natural logarithm of both sides...

[ewmayer]

Originally Posted by sagan_fan

I have a feeling that the number 19683^(387420489)-2 is either a square or on the form a^a. Anybody wanna help me prove it?

N = 19683^387420489 - 2 has the small factor f = 105049853 .
This factor is not repeated, so N is definitely not a prime power of any kind,
though I suppose it could contain the (foo)th power of some prime (bar)
somewhere in its complete factorization. But that's all
very vague. Give us some mathematical reason to believe it's an
interesting number of any sort, and maybe we'll spend more time
looking at its (so far completely unremarkable) properties.

Ooh, look, by adding 4 to it I get an even BIGGER number. Amazing.