Solving A Diophantine Equation

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[XinLi]

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.

[jisuuqm]

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

[vladimirkaplun]

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

[nasrullahhusnan]

(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}

[devondevon]

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

[Prab]

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)

[chaosredefined]

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.

[scottleung]

I also got 0 and 2 as the only solutions.

[RealPersistences]

Anyone know what to do when the b and c values in a factoring equation are the same, as in 2x^2 - 9x -9?

[bkkboy-cmeb]

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

[wannabeactuary]

Let f(m) = 3^m/m^3
ln f = m ln 3 - 3 ln m
f'/f = ln 3 - 3/m
f' > 0, if m >= 3