A Nice Diophantine Equation

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a²b³ = 6^6 = (2² 3²)³
a² = (2² 3² / b)³
As only integer solutions are admitted, the right side of the equality must be a perfect square, since the left side is equal to a². Then:
b = 1 ==> a² = (2² 3²)³ ==> a = ±(2 . 3)³ = ± 216
b = 2² ==> a² = (3²)³ ==> a = ±3³ = ± 27
b = 3² ==> a² = (2²)³ ==> a = ±2³ = ± 8
b = 2² 3² ==> a² = 1³ ==> a = ±1³ = ± 1
All solutions: (-216, 1) ; (216, 1) ; (-27, 4) ; (27, 4) ; (-8, 9) ; (8, 9) ; (-1, 36) ; (1, 36)

walterufsc

You have done math proud by including complex solutions in many of these problems. Can you do a series that includes quaternion solutions to many of these problems?

I recognize that this particular one is limited to integers.

deltalima

My method, raise to power 1/6 both sides --->
a^(1/3) * b^(1/2) = 6 ---> b^(1/2) must be divisor of 6 ---> b^(1/2) = 1, 2, 3, 6 ----> b = 1, 4, 9, 36
substitute and you have corresponding values of a

WahranRai

Very nice solutions. Eight in all, neatly found.

dentonyoung

a^2 * b^3 = 6^6. Now raise both sides to the power of 1/6. cbrt(a) * sqrt(b) = 6. I'll divide both sides by cbrt(a) and then square both sides to get: b = [6 / cbrt(a)]^2. Notice that b has to be an integer, so the RHS is also an integer. Therefore cbrt(a) has to be a divisor of 6. This gives us cbrt(a) = +-1, +-2, +-3, +-6. Therefore a = +-1, +-8, +-27, +-216. And if you put these into the formula for b, we get b = 36, 9, 4, 1. Therefore the solutions are: (a, b) = {(1, 36), (-1, 36), (8, 9), (-8, 9), (27, 4), (-27, 4), (216, 1), (-216, 1)}.

DrQuatsch

I found it simpler just to note that 6^6 = 2^6 * 3^6 OR 1^6 * 6^6. Either way, we have two factors on the left and two factors on the right. So, just set them equal to each other and find all the combinations. E.g., let a^2 = 2^6. Solve for a. Then b^3 = 3^6. Solve for b. Then let a^2 = 3^6. Etc. It's pretty quick, and doesn't even require paper.

j.r.