A simple math problem?

Question

Mathematics

Here is a simple math problem for you.
It is necessary to prove that for any natural m, n > 1, the following inequality holds:
3^m + 1 <> 2^n
Instead of 3, in the general case, there can be any prime number.
I will post the answer tomorrow if anyone is interested. Solution in several lines without any tricks.

Answer 1

(2^n mod 8) = 0 when (3^m+1) mod 8 c {2,4}
* throw away change by hand

Answer 2

42

Answer 3

3^1+1=2^2?

Answer 4

You can prove it by induction. If this inequality is true for some natural k, then it is only required to prove its correctness for k + 1
And so, suppose that the inequality is true for some m. Then for m + 1 we get 3*3^m <> 2^n - 1. If 3^m were equal to 2^n - 1, then we would get 3 == 1. But this is not so

Answer 5

the first thing that came to mind was:
3^m+1 <> 2^n
=>
3^m <> 2^n-1
2^n-1 is a prime number (see Mersen numbers), and 3^m is obviously no.

Answer 6

Proof of the general case p^m + 1 <> 2^n
1) Assume that m is odd. Then, after expanding the polynomial on the left side, one of the factors will turn out to be an odd number (not necessarily a prime number). For any n, this number cannot appear on the right side.
2) Assume that m is even.
2.1) Consider the case of even n. After transferring p^m to the right side and factoring it out, on the right side we get (2^(n/2) - p^(m/2)) * (2^(n/2) + p^(m/2) ), which obviously cannot be equal to one on the left side.
2.2) Consider the case of odd n. Let's add 2 * p^(m/2) to the left and right sides. On the left side you get a square (p^(m/2) + 1)^2, and on the right side 2 * (2^(n-1) + p^(m/2)), that is, the left side is divided by a square deuce, and the right, only 2.

Answer 7

The right hand side is always even.
The left side is always odd, because:
3^m+1=(2+1)^m+1 When expanded in terms of Newton's binomial, we obtain the sum of a power of two multiplied by the binomial coefficient. those. even number. add one to it and get odd.

Answer 8

My solution was probably the same as suggested above.
In binary, the last 4 digits for 3^m will be repeated:
0011
1001
1011
0001
Obviously adding one will never give all zeros. Therefore, it will never be equal to 2 ^ n for any m and n > 1.
Similar patterns exist for all prime numbers, and not just for 3.
For those cases where the patterns are not obvious, you can start from the fact that the binary notation any prime number to any power greater than 1 is 0. When multiplying such a number by itself, 0 will never get rid of this. And in order for the result to become equal to a power of two when 1 is added, all digits must be one, which is not.