A Nice Diophantine Equation | 3ˣ - x³ = 1

[@misterdubity3073]

There is also a non-integer solution at x = -0.84584 approx. Nice graph on Desmos 3^x vs x^3 + 1

[@ubncgexam]

There is also a 4th solution at x ≈ 3.2206447... 😉

[@misterdubity3073]

@@ubncgexam Good pick up. I didn't see it because I graphed what I graphed. Graphing 3^x - x^3 - 1 shows the zeros nicely.

[@MrGeorge1896]

Exponential growth beats polynomial growth so we just have to examine x = 0, 1, 2 and 3 and show that for x >=4 the LHS is strictly increasing. Negative values for x can be rejected as we would get fractions on the LHS.

[@premkumarsr4021]

Beautiful. No words to express my happiness

[@SyberMath]

Glad to hear that! 🥰🧡

[@andypandy6063]

This one was really fun.

[@MR..mohamedelsayed]

The equation has 4 solutions if we follow the analytical solution. We notice that the (3^x) exponential function intersects with (x^3+1) the real function. There are 4 points, two of which are easily accessible either by trial or by methods that we might consider algebraic, which are {0, 2}. However, there are two other solutions. Can we reach them through algebraic methods and not graphical? This is the topic of research and discussion. Thank you to the channel owner for the ideas presented.

[@scottleung9587]

I also got both 0 and 2, but I just plugged them in.

[@SyberMath]

Quoting John Scott: "Scott Leung's solution is best. Hands down. Always try this first when integers are required."

[@johnscott3484]

Scott Leung's solution is best. Hands down. Always try this first when integers are required.

[@SyberMath]

😉😁

[@scottleung9587]

Thanks so much!

[@nasrullahhusnan2289]

x=2 as (3^x)-x³=3²-2³ =9-8=1

[@pedrovargas2181]

x = {0,2}, by try-and-fail. Third root, no idea. 8:40. a = 0.

[@goldfing5898]

9:32 The solution x = 0 can easily be found by guessing, also the solution x = 2. I would use graphs of the functions to watch for potential other solutions, and maybe some calculus.

[@tontonbeber4555]

It's more interesting to solve it in reals ... 4 solutions :))

[@SG49478]

I have an issue with this method. Transforming an equation into a system of equations is not an equivalent transformation and has the potential for losing solutions. Your method finds 2 solutions, but lacks proof that there is no third one. Luckily there are no other integer solutions because it is relatively easy to proof that for x>2 the difference of 3^-x^3 is consistently increasing. To come up with 0 and 2 as solutions for x there are less complicated ways to do that. I like the approach of Random Jin, which is more straight forward and also includes proof, that there are no other solutions.

[@SyberMath]

Factoring shows there are no others

[@SG49478]

@@SyberMath I can't really follow you. Where did you factor the equation 3^b+3^a+1=3^2a+3? The equation itself is a sum.

[@jpbobinus1377]

Please, why is this equation diophantine?

[@SyberMath]

We are only interested in integer solutions

[@msmbpc24]

X=0 or X=2

[@zawatsky]

До сих пор не понимаю, зачем расписывать решение уравнения, которое решается простой подстановкой.

[@-rahul-2908]

huh? what's the substitution

[@zawatsky]

@@-rahul-2908 ну, я просто сначала подставил 3, а когда не получилось - 2. )

[@-rahul-2908]

@@zawatsky Ah, okay - хорошо

[@vladimirkaplun5774]

As at x=4 3^4=81 while 4^3 is just 64 and (x+1)^/x^3 is definitely less than 4 only x=0,1,2,3 should be tested. No number theory, nothing

[@SyberMath]

Nice!

[@kianmath71]

X =2

[@PROTAEQUESO98]

x = 2

[@walterwen2975]

A Nice Diophantine Equation: 3^x - x^3 = 1; Find the integer solutions of x 3^x - x^3 = 1; 3^x > x^3, x ≠ 1, 3 > x ≥ 0 First method: When: x = 0; 3^x - x^3 = 3^0 - 0 = 1; Confirmed When: x = 2; 3^2 - 2^3 = 9 - 8 = 1; Confirmed Second method: Let: x = 2y 3^x - x^3 = 3^2y - (2y)^3 = (3^2)^y - (2^3)(y^3) = 9^y - 8(y^3) = 1 When: y = 0, x = 2y = 0; 3^x - x^3 = 9^y - 8(y^3) = 9^0 - 8(0) = 1; Confirmed When: y = 1, x = 2y = 2; 3^x - x^3 = 9^1 - 8(1^3) = 9 - 8 = 1; Confirmed Final answer: x = 0 or x = 2

[@jmlfa]

I don't get it. It took me two seconds to figure out that x=0.

[@MushookieMan]

But that's only one solution...

[@GsBabu-sk6iv]

Roger is bad at math.

[@Allstuffthatfascinates]

Today I did a proof of geometry on my own. Theorem: In every triangle, at least one altitude from one of the vertices lies inside it. Proof by me: Let ΔABC be a triangle that has all it’s altitudes lying outside it. We know that if an angle of a triangle is obtuse, then altitudes from other two vertices lie outside the triangle. Now since all altitudes of ΔABC lie outside, then at least two of it’s angles must be obtuse. So, let ∠A and ∠B be obtuse. ⇒ ∠A+∠B>90∘+90∘ ⇒∠A+∠B>180∘ ⇒∠A+∠B+∠C>180∘ But, ∠A+∠B+∠C=180∘ So, we arrive at a contradiction and our assumption was wrong. ⇒ No two angles in a triangle can be obtuse and at least one altitude of every triangle lies inside it. Hence Proved. Thanks.

[@ubncgexam]

And I prooved the Rieman's hypothesis... But as once Fermat stated "the margin is too small to write it down"😂😂😂... WTH you are proving? 😮🙄🙄

[@rakenzarnsworld2]

x = 2