Current Events > Given that (2^x) = (5^y) + 1 where x and y are integers...

(+) Yeah! (+)

How many different solution pairs of x & y exist and why? Are there an infinite number of them?

--
---

Gotta use the Log commands

Search for shit on youtube about whatever the title of the chapter is or w.e
---

No solutions exist since every 2^n-1 that is divisible by 5 is also divisible by 3 and thus can't be a power of 5. Basic number theory.

If a solution existed, then infinite solutions would exist due to the property of composite Mersenne numbers, but no solution exists.
---

I never said x and y have to be positive. x=1, y=0 is one solution. But is it the only one?

--
---

FLOUR posted...

I never said x and y have to be positive. x=1, y=0 is one solution. But is it the only one?

--

Yes as, except for the case you provided...
ZeldaMutant posted...

No solutions exist since every 2^n-1 that is divisible by 5 is also divisible by 3 and thus can't be a power of 5. Basic number theory.

---

OK, how about positive integer solutions of (3^x)=(2^y)+1? That should be a tad more difficult.

--
---

Yes there is an infinite number.
---

for negative x and y I don't think there can ever be solutions since 2 and 5 are prime there will always be a fraction value for 2^x-5^y. But for x,y>=0, 5^y+1 always end in a 6 and 2^x only ends in a 6 if x=y+4 (x has to be bigger than y) excluding the special case of x,y=0

2^(y+4)=5^y+1
2^4*(2^y)=5^y+1^y
2^4=(5^y+1^y)/(2^y)
2^4=(5/2)^y+(1/2)^y

5/2>2 so this will never be true if y>4

I don't know what I'm doing...

y must be of a form 6k+3 (other than x=1,y=1) due to 2^y having to be equal to (8 modulus 9)
for x>4 y must be of a form 54k+27
solutions that I've found so far
x=1 y=1
x=2 y=3
given the complexity that y's form increases, I would say that is all of the solutions
---

I just realized I screwed up. I said x=y+4, but I think its more appropriate to say x=4my, so x is any number bigger than y that is divisible by 4 that way 2^x ends in a 6. I still can't prove it though :(