An exponential Diophantine equation

Using number theory tricks to solve an exponential Diophantine equation.

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[braindead]

For all the people commenting z=5, that’s not where the mistake is. The equation he should’ve written was 4=2^{2c} which gives you c=1 and z=3 (which is the correct answer). He instead wrote 4=2^c and got c=2 which is wrong. Try plugging in z=5 and see what happens.

[Bekonisko]

If c=2, shouldn't z=5?
Also, why is (2a+1)^b-1 divisible by a?

[jkid]

I like the way you chop away at the space. Very intuitive

[mrfork]

Around 10:50, writing (2a + 1)^b - 1 = 2a^(2c+1) is valid only if a>2
Following the Mickael logic we obtain a = 1 which is then a contradiction. So a =2 or 1:
The case a= 2 gives (5^b - 1)(5^b + 1) = 4(16^c). 5^b - 1^b = 4d, 5^b + 1 = 4d + 2 . We obtain 2d(2d+1) = 16^c only possible if 2d + 1 = 1 => d = 0 => b = 0 contradiction
Then a = 1

[khoozu]

I think we have to prove why a^(2c+1) must belong to the factors of (2a+1)^b-1 only. We know (2a+1)^b-1=na, then (2a+1)^b+1=na+2=2(mod a)
(2a+1)^b+1 is divisible by a if and only if a=1 or 2. But if a=1, then the place of a^(2c+1) become not important; if a=2, the min value of (2a)^(2c+1)=4^3=2^6. For both gcd to be 2, only one solution can be true➡️(2a+1)^b-1=2, (2a+1)^b+1=2^5
But now their difference is greater than 2, which contradicts with the fact of [(2a+1)^b+1]-[(2a+1)^b-1]=2

[weonlygoupfromhere]

It's so funny lol. I understand calc and all that hard crap in a few goes, but number theory confuzzles me

[reeeeeplease]

Cant we just use the fact(?) that
x^n = y^m + 1
with x, y > 0 and n, m > 1
is only solved by
3^2 = 2^3 + 1

Im not quite sure if that is the correct statement but i think there was something quite similiar

[benterrell]

Nice. I need to think a bit more about why even numbers separated by 2 have a GCD of 2. It will come to me with the morning coffee. I had a seven hour drive yesterday.

[alomirk]

You can solve this question extremely quickly by moving the x^z to the other side and then using catalans conjecture

[Blabla]

at 11:19, you stick all factors of 2 except one into the second term. Why is that?

[julienmallet]

i dont understand why at the beginning
x^z + 1 ≡ 0 [mod x+1]

[s]

Wouldn't by inspection from [Mihăilescu's theorem / Catalan's conjecture] the only solution is 3^2 = 2^3 +1? That kind of skip everything inbetween 😅

[RalphDratman]

That is amazing! Eighth-grade math done with such grasp and dexterity that what seems like an impossible problem is solved very, very quickly.
Bravo

[amidhmi]

Can someone please explain the part at 10:25? I dont understand why he factored it that way. Maybe some of the 2 of the right factor can go to the left factor?

[BarryRowlingsonBaz]

A better place to stop would be to put those values into the original equation and check they work :)

[wavyblade]

10:28 how do we know they have the gcd of 2? What if a=2?

[brankojangelovski]

You can also finish easily with zsigmondy

[andrewlayton]

Interesting in that it devolves into the Catalan Conjecture.
After finding the difference of perfect squares, we know that the factors on the left side are even and have difference of 2. Wouldn't that quickly lead us to the idea that the factors are 4 and 2?

[Denosophem]

Hahaha I actually like this a lot tbh!

[Denosophem]

I’d think it’s pretty close but I don’t know math