Solving A Diophantine Equation

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18 сен 2024

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Комментарии : 34

@XinLi 20 дней назад

Why go through such a long winded process? 1) First, note that 3^x grows much faster than a cubic function. Thus, for large ms, 3^m outstrips m^3. 2) When m = 3, 3^m and m^3 are equal. Beyond that, the two functions will never touch again. 3) Similarly, for negative integers, 3^m rapidly approaches zero, while m^3 gets to be a super negative number. 4) Thus, you only have to try integer values of 0,1,2 to find a solution. Takes all of 30 seconds.

@SyberMath 20 дней назад

Nice!

@mrityunjaykumar4202 15 дней назад

it would have been better if he would have proved by induction that for m>=4; 3^m-m^3>1

@abdulhusseinalsultani9222 7 дней назад

1, 2 and 3 are small numbers so by little inspection we can say m=2 3^2_2^3=9_8=1 answer

@user-ji5su2uq9m 18 дней назад

from 6:40 3^k = 3^(2n) - 3^(n+1) + 3 if k = 0 , 3^(2n) - 3^(n+1) + 3 = 1 => (3^n - 1)(3^n - 2) = 0 => n = 0 => m =0 if k ≥ 1 , 3^(k - 1) = 3^(2n - 1) - 3^n + 1 => 3^(k - 1) ≡ 1 (mod 3) => k - 1 = 0 => n =1 => m =2

@nasrullahhusnan2289 18 дней назад

(3^m)-m³=1 The trivial solutiin is m=0 For other possible silution, take modulo 2: mod(3^m,2)-mod(m³,2)=1 [mod(3,2)]^m-[mod(m,2)]³=1 (1^m)-1=[mod(m,2)]³ mod(m,2)=0 --> m=2k, k an integer 3^m=m³+1 --> 3^m=(m+1)(m²-n+1) (m+1) | 3^m (2k+1) | 9^k --> k=1 m=2 Therefore m={0, 2}

@vladimirkaplun5774 20 дней назад

When m is 3 3^m=m^3. Starting from m=4 3^m will be much bigger than m^3->81>64 . Hence just try m= 0,1,2 and that's it

@hakanerci4372 19 дней назад

0 and 2 is ok. But 1 not good

@SyberMath 19 дней назад

good thinking!

@devondevon4366 14 дней назад

Answer 2 and 0 3^m - m^3 =1 3^ m= m^3 + 1 3^integer is always odd Hence, 3^ m is odd Hence, m^3 + 1 is odd Hence, m^3 is even Hence, m is even If m is even, then 3^ m unit digit is either 1 or 9 the difference between 3^ m and m^3 =1 3^ 0 = 1 0^3 =0 and 3^ 2 =9 and 2^3 =9 So 0 and 2 are two solutions since the difference between 3^m and m^3 is 1 in both cases. Let's try some more 3^4 and 4^3 81 and 64, a difference of 17 ' 3^6 and 6^ 3 729 and 216, a difference of 513 The difference becomes larger when m increase Hence, the solutions are 0 and 2 When m is negative, the difference 3^m is not an integer, but m^3 is

@Prab2212 19 дней назад

Hi SyberMath, big fan. I had a similar solution to you: 3^m = (m+1)(m^2-m+1) the terms on the RHS must be powers of 3, since they multiply to make a power of 3 this means that one of these terms must be a factor of the other term (m^2-m+1)/(m+1) = (m-2) + 3/(m+1) because (m+1) divides (m^2-m+1), 3/(m+1) has to be an integer hence (m+1) is a factor of 3 hence m= 0 or 2 (I am a high school student and just also wanted to thank you for improving my problem solving ability allowing me to qualify for a number of mathematics competitions through my school)

@SyberMath 19 дней назад

Wow! Thank you. I'm glad to hear that

@chaosredefined3834 20 дней назад

3^m - m^3 = 1 3^m = m^3 + 1 3^m = (m+1)(m^2 - m + 1) m+1 | 3^m Hence m+1 is a power of 3. Case 1: m+1 = 3^0 = 1 Hence m = 0. So, 3^0 - 0^3 = 1 - 0 = 1. Valid solution. Case 2: m+1 > 1. This means that m+1 is a power of 3, so m = 3k+2 But also, m^2 - m + 1 is a power of 3. And that's 9k^2 + 12k + 4 - 3k - 2 + 1 = 9k^2 + 9k + 3. Well, taking out a multiple of 3, we get 3(3k^2 + 3k + 1). So, 3k^2 + 3k + 1 is a power of 3. But it is clearly not divisible by 3, so it must be 1. Therefore, k = 0, and m = 3(0) + 2 = 2. Testing this, we get 3^2 - 2^3 = 9 - 8 = 1, so it's valid. Solutions are m=0 and m=2.

@SyberMath 20 дней назад

Very good!

@jamesmarshall7756 20 дней назад

Great solution !

@SyberMath 20 дней назад

Thanks!

@scottleung9587 20 дней назад

I also got 0 and 2 as the only solutions.

@yusufdenli9363 20 дней назад

Nice solution

@SyberMath 20 дней назад

Thanks!

@Nikioko 20 дней назад

m = 2

@rakenzarnsworld2 20 дней назад

m = 0 or 2

@dariosilva85 20 дней назад

What are you if not a mathematician?

@SyberMath 20 дней назад

Someone who likes math! 😜

@dariosilva85 20 дней назад

@@SyberMath You gotta be at least engineer, right?

@scottcowan8036 17 дней назад

To me, at least, a mathematician is someone who works to develop entirely new ideas and results in mathematics, whereas a math educator is someone who takes existing ideas and results and teaches/explains them (and how to play around with them to problem-solve) to others. You can be a math educator without being a mathematician, or you can be a mathematician without being a math educator, or you can be both. I would say that SyberMath is a math educator. (As am I.)

@bkkboy-cm3eb 20 дней назад

3^m-m³=1 m=0 → ok. when m>0, -m³≡1 (mod3) → m=3k-1 (k>0) 3^m=m³+1=(m+1)(m²-m+1) =3k(9k²-9k+3)=9k(3k²-3k+1) ∴3^p=9k ···① 3^(m-p)=3k²-3k+1 ···② ②→ m-p=0 → 3k(k-1)=0 → k=1 ①→ 3^p=3^m=9 → m=2 ∴m=0, 2