Mixing
Primes of the form $a^n-(a-1)^n$
$begingroup$
For a given $n$, consider the assertion:
$exists a in Bbb Z : a^n-(a-1)^n text{is prime}tag*{}$
How can one do one of the following:
- Prove that the assertion is true for all integer $n > 1$
- Prove that the assertion is false for all integer $n > 1$
- Characterize the values of $n > 1$ for which the assertion is true
The assertion is clearly true (with $a=2$) for all values of $n$ that are Mersenne prime exponents. Also, we only need to consider positive values for $a$. If $a<0$ then $a^n-(a-1)^n$ is negative (and hence not prime) when $n$ is even and is equal to $(1-a)^n-((1-a)-1)^n$ when $n$ is odd.
I suspect the assertion is true for all $n > 1$, but I have no idea how to prove it.
EDIT (thanks to @Malcolm):
The assertion is false whenever $n$ is composite. This is because $a^{uv} - (a-1)^{uv}$ is divisible by $a^u - (a-1)^u$ and by $a^v - (a-1)^v$. (This can be shown by induction on $v$ for $a^u - (a-1)^u$ and vice versa.)
I'd still like to know if the assertion is true for $n$ prime and not a Mersenne prime exponent.
EDIT 2:
It turns out that this family of numbers has been studied before, although not extensively, as one form of generalized Mersenne numbers. (See, for instance, the 2003 conference paper Generalized Mersenne prime numbers: characterization and distributions by Vladimir Pletser.) It also turns out that there are indeed primes of the form $a^n - (a-1)^n$ for values of $n$ that are not Mersenne prime exponents. For instance,
$M_{11} = 2^{11} - 1 = 2047 = 23cdot89tag*{}$
while
$6^{11} - 5^{11} = 313968931tag*{}$
is prime. Pletser established by brute force that there are primes of the form $a^n - (a-1)^n$ for all prime $n$ through 31. He and T. D. Noe have tabulated the smallest such prime for the first 80 primes. See OEIS A121620.
prime-numbers
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
$endgroup$
|
show 6 more comments
$begingroup$
For a given $n$, consider the assertion:
$exists a in Bbb Z : a^n-(a-1)^n text{is prime}tag*{}$
How can one do one of the following:
- Prove that the assertion is true for all integer $n > 1$
- Prove that the assertion is false for all integer $n > 1$
- Characterize the values of $n > 1$ for which the assertion is true
The assertion is clearly true (with $a=2$) for all values of $n$ that are Mersenne prime exponents. Also, we only need to consider positive values for $a$. If $a<0$ then $a^n-(a-1)^n$ is negative (and hence not prime) when $n$ is even and is equal to $(1-a)^n-((1-a)-1)^n$ when $n$ is odd.
I suspect the assertion is true for all $n > 1$, but I have no idea how to prove it.
EDIT (thanks to @Malcolm):
The assertion is false whenever $n$ is composite. This is because $a^{uv} - (a-1)^{uv}$ is divisible by $a^u - (a-1)^u$ and by $a^v - (a-1)^v$. (This can be shown by induction on $v$ for $a^u - (a-1)^u$ and vice versa.)
I'd still like to know if the assertion is true for $n$ prime and not a Mersenne prime exponent.
EDIT 2:
It turns out that this family of numbers has been studied before, although not extensively, as one form of generalized Mersenne numbers. (See, for instance, the 2003 conference paper Generalized Mersenne prime numbers: characterization and distributions by Vladimir Pletser.) It also turns out that there are indeed primes of the form $a^n - (a-1)^n$ for values of $n$ that are not Mersenne prime exponents. For instance,
$M_{11} = 2^{11} - 1 = 2047 = 23cdot89tag*{}$
while
$6^{11} - 5^{11} = 313968931tag*{}$
is prime. Pletser established by brute force that there are primes of the form $a^n - (a-1)^n$ for all prime $n$ through 31. He and T. D. Noe have tabulated the smallest such prime for the first 80 primes. See OEIS A121620.
prime-numbers
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
$endgroup$
$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
1
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
2
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
1
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22
|
show 6 more comments
$begingroup$
For a given $n$, consider the assertion:
$exists a in Bbb Z : a^n-(a-1)^n text{is prime}tag*{}$
How can one do one of the following:
- Prove that the assertion is true for all integer $n > 1$
- Prove that the assertion is false for all integer $n > 1$
- Characterize the values of $n > 1$ for which the assertion is true
The assertion is clearly true (with $a=2$) for all values of $n$ that are Mersenne prime exponents. Also, we only need to consider positive values for $a$. If $a<0$ then $a^n-(a-1)^n$ is negative (and hence not prime) when $n$ is even and is equal to $(1-a)^n-((1-a)-1)^n$ when $n$ is odd.
I suspect the assertion is true for all $n > 1$, but I have no idea how to prove it.
EDIT (thanks to @Malcolm):
The assertion is false whenever $n$ is composite. This is because $a^{uv} - (a-1)^{uv}$ is divisible by $a^u - (a-1)^u$ and by $a^v - (a-1)^v$. (This can be shown by induction on $v$ for $a^u - (a-1)^u$ and vice versa.)
I'd still like to know if the assertion is true for $n$ prime and not a Mersenne prime exponent.
EDIT 2:
It turns out that this family of numbers has been studied before, although not extensively, as one form of generalized Mersenne numbers. (See, for instance, the 2003 conference paper Generalized Mersenne prime numbers: characterization and distributions by Vladimir Pletser.) It also turns out that there are indeed primes of the form $a^n - (a-1)^n$ for values of $n$ that are not Mersenne prime exponents. For instance,
$M_{11} = 2^{11} - 1 = 2047 = 23cdot89tag*{}$
while
$6^{11} - 5^{11} = 313968931tag*{}$
is prime. Pletser established by brute force that there are primes of the form $a^n - (a-1)^n$ for all prime $n$ through 31. He and T. D. Noe have tabulated the smallest such prime for the first 80 primes. See OEIS A121620.
prime-numbers
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
$endgroup$
For a given $n$, consider the assertion:
$exists a in Bbb Z : a^n-(a-1)^n text{is prime}tag*{}$
How can one do one of the following:
- Prove that the assertion is true for all integer $n > 1$
- Prove that the assertion is false for all integer $n > 1$
- Characterize the values of $n > 1$ for which the assertion is true
The assertion is clearly true (with $a=2$) for all values of $n$ that are Mersenne prime exponents. Also, we only need to consider positive values for $a$. If $a<0$ then $a^n-(a-1)^n$ is negative (and hence not prime) when $n$ is even and is equal to $(1-a)^n-((1-a)-1)^n$ when $n$ is odd.
I suspect the assertion is true for all $n > 1$, but I have no idea how to prove it.
EDIT (thanks to @Malcolm):
The assertion is false whenever $n$ is composite. This is because $a^{uv} - (a-1)^{uv}$ is divisible by $a^u - (a-1)^u$ and by $a^v - (a-1)^v$. (This can be shown by induction on $v$ for $a^u - (a-1)^u$ and vice versa.)
I'd still like to know if the assertion is true for $n$ prime and not a Mersenne prime exponent.
EDIT 2:
It turns out that this family of numbers has been studied before, although not extensively, as one form of generalized Mersenne numbers. (See, for instance, the 2003 conference paper Generalized Mersenne prime numbers: characterization and distributions by Vladimir Pletser.) It also turns out that there are indeed primes of the form $a^n - (a-1)^n$ for values of $n$ that are not Mersenne prime exponents. For instance,
$M_{11} = 2^{11} - 1 = 2047 = 23cdot89tag*{}$
while
$6^{11} - 5^{11} = 313968931tag*{}$
is prime. Pletser established by brute force that there are primes of the form $a^n - (a-1)^n$ for all prime $n$ through 31. He and T. D. Noe have tabulated the smallest such prime for the first 80 primes. See OEIS A121620.
prime-numbers
prime-numbers
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
edited Jan 17 at 21:48
Ted Hopp
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
asked Jan 16 at 18:35
Ted HoppTed Hopp
1607
1607
$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
1
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
2
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
1
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22
|
show 6 more comments
$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
1
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
2
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
1
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22
$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
1
1
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
2
2
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
1
1
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22
|
show 6 more comments
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$begingroup$
To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition.
$endgroup$
– pwerth
Jan 16 at 18:42
$begingroup$
@pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that.
$endgroup$
– Ted Hopp
Jan 16 at 18:52
1
$begingroup$
The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$
$endgroup$
– Malcolm
Jan 16 at 20:26
2
$begingroup$
@Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$.
$endgroup$
– Ted Hopp
Jan 16 at 21:04
1
$begingroup$
A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime.
$endgroup$
– Cheerful Parsnip
Jan 16 at 21:22