Solving 3^x+4^x=5^x

Solving this Pythagorean exponential equation 3^x+4^x=5^x

#shorts #pythagorean #algebra

[tylerle]

isn't it an always decreasing function? since a^x → 0 as x → ∞ for 0 < a < 1? the point made in the video still stands (the expression the left side will only be equal to 1 at one point) but yeah

[reidpattis]

Classic example of finding A SINGULAR solution by guess/intuition and proving the uniqueness of the solution. Pretty satisfying stuff.

[ms_slytherin]

That's the most popular Pythagorean triple, so 2 came in my mind the moment I saw the equation 😅

[Ninja]

I’m actually curios, but what about complex solutions?

[rageprod]

This equation is of the type a^x + b^x = c^x where a, b and c are positive integers. We know that when x=1 and x=0 the equation is false and when x=2 it is true. We know by Fermat's Last Theorem that no triple of integers satisfy this equation for x>2, so 2 must be the only natural solution.

(This tells you nothing about negative or non-integer solutions, let alone complex ones, but still interesting imo)

[elosant]

A trivial proof would be to show that elliptic curves over the rationals can be obtained by a modular form.

[gurkiratsinghtha]

This is a special case of the fermat's last theorem

[metalheadmaniac]

I dont know why I watch this channel I don't even like math the only reason I use it is because I'm a game programmer
there is just something about this channel that I can't get enough of

[vyankateshwarshendye]

3/5 and 4/5 are sins and cosines of something 37... so they add up when squared to unity

[orisphera]

Here are two other solutions:
Move everything to the left part (i. e. subtract the right part) and take the derivative. I think it's easy to finish this solution.
According to the description, this is a Pythagorean equation, so x should be an integer. Case x>2 is a corollary of the Fermat theorem, x<-2 is a slightly more difficult one but also not very hard, and the rest can be considered.

[benamiel]

We dont have to do all that. By fermat last theorem the only equation of x^z + y^z = n^z is when z <= 2

[chrisng]

or just use a theorm with a magnificent proof, which this comment is too small to contain

[AvNews_Info]

I knew it because of the 3, 4, 5 triangle

[dariushdarouie]

If we consider sin x=0.6 so sin x^2= 0.36 & cos x^2 = 1 - sin x^2 = 0.64 so sin x = 0.8 so we can result x=2

[firebird]

You’re proving Fermat’s last Theorem, good luck, i have a really neat proof for this but it won’t fit in this comments section.

[atzuras]

by Fermat's Last Theorem there cannot be an integer solution for x > 2.
Which, in my mind, also negates any other real result.
I have a nice proof but the comment section is too small to write it down over here.

[thalfie]

The reason being why this is only 2 is also a connection with the proof of Fermat's Last Theorem, where there are no possible numbers for n (or x in this case) where it could fit the equality statement. Try as you might, it never works.

[codelif]

Another answer is Fermat's Last Theorem. Hence, 2

[greatorionbelt]

Wheres the "And then we are done!" :(((

[koktonglan]

Solve this by using AM-GM inequality if both power are same.