Find all (x,y=integer number) so that
x^3 - 3^y = 100

Question

Find all (x,y=integer number) so that
x^3 - 3^y = 100

anonymous · Answer

seems unlikely
do you have a solution?

anonymous · Answer

Possible typo?

anonymous · Answer

i hope so

raden · Answer

one of solution i got x=7, y=7

anonymous · Answer

otherwise you get $y=\frac{\ln(x^3-100)}{\ln(3)}$ and

raden · Answer

sorry, x=7, y=5

anonymous · Answer

\(7^3-3^7=-1844

anonymous · Answer

i'll be damned

experimentx · Answer

x^3  = 100 + 3^y <-- exponential
^ cubic

anonymous · Answer

I'd probably go brute force on this one.
*opens spreadsheet software*

raden · Answer

experimentX, yeps... exponential
any one help me find all solutions, and how?
i got the answer with trial and error, very tired for me...

experimentx · Answer

hold on .. they are different variables.

raden · Answer

yea, it's the problem...

anonymous · Answer

You can use the formula satellite derived and a graphing calculator or computer spreadsheet and do trial and error for numerous integer values of x and see which y values it returns are integers.

anonymous · Answer

The next one I found was (81,12) and I have a couple more after that.

anonymous · Answer

There appears to be a pattern. My conjecture is that besides the (7,5), the y values are multiples of 3.

anonymous · Answer

Furthermore, I would conjecture that the x's are integer powers of 3.

anonymous · Answer

is there one with $x=27$ ?

anonymous · Answer

Hmmm, I seem to be getting round-off error, though . . .

anonymous · Answer

No, so far I have (7,5), (81,12), (243,15), (729, 18), (2187, 21), but after that things are looking fishy.

anonymous · Answer

Going to try to up the precision...

experimentx · Answer

(81,12) = 0

anonymous · Answer

i am trying to think what conditions make $x^3+100$  a power of 3

anonymous · Answer

rather $x^3-100$

anonymous · Answer

Yep, I'm getting serious round-off error here. Drat.

anonymous · Answer

is it factorable : x^3+100 ?

anonymous · Answer

243, 15 doesn't work

raden · Answer

i think can't be factorise it mukushla...

anonymous · Answer

@CliffSedge   you are getting 0 for each, not 100

experimentx · Answer

they are all zero ...

anonymous · Answer

Yep, I'm using more powerful tool now. So far I've tested all values up to x=2600 and haven't found any others.

raden · Answer

so, just one solution like i got?

anonymous · Answer

Good possibility. I'm up to x=6000 now and still haven't found any others.

raden · Answer

Ok, thanks all for ur respons
dont forget, give m a medals, hahaha....

anonymous · Answer

Interesting little equation you found there. I'm going to stop looking. I'm up to x=10,000 now and still no others.

experimentx · Answer

this is a one hell of a question ... i googled.

experimentx · Answer

probably not my stufff there is a link that might lead you to solution http://math.stackexchange.com/questions/43326/on-natural-solutions-of-the-equation-y3-3x-100

raden · Answer

Ok, thank you verymuch experimentX...

experimentx · Answer

no probs at all

anonymous · Answer

tough one

experimentx · Answer

up to you man!!

experimentx · Answer

maybe someday i'll try to get number theory

anonymous · Answer

I think I found some other ones. I'm double-checking them now. I might be beyond the limit of precision of my computer. *eek*

anonymous · Answer

Dang it, I think it's still just round-off error again...  Grr. Need a better processor than what I have to try brute force methods. I thought I found one around x=531441, but I'm getting x^3 - 3^y = 0 again.