$n in mathbb N$, find all values of $n$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number. [duplicate]
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This question already has an answer here:
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
3 answers
Find all values of $n in mathbb N$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.
Let $k=3^{2n+1}-2^{2n+1}-6^n$. By testing with numbers I got that $k$ cannot be prime for any value of $n$ (tested to $n=20$, so I assumed it's same for every value of $n$).
For $n=1$, $k<0$ $implies$ $k$ is not prime. (EDIT: oh... it's actually prime if $n=1$. Stupid me calculated $3^3=9$)
For $n>1$:
I managed to prove that if $n$ is even, $k$ is not a prime:
If $n$ is even, then $$3^{2n+1}equiv3 (mod 10),quad 2^{2n+1}equiv2 (mod 10),quad 6^nequiv6 (mod 10)$$
$$implies kequiv5 (mod 10)$$ So $k$ is divisible by 5 and therefore not prime.
Using the same method for if $n$ is odd, I get: $kequiv3 (mod 10)$, which doesn't prove anything.
How do I solve for when $n$ is odd?
elementary-number-theory prime-numbers modular-arithmetic
asked Dec 19 '18 at 20:16
PeroPero
1497
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marked as duplicate by Bill Dubuque modular-arithmetic
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Dec 19 '18 at 20:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
3 answers
Find all values of $n in mathbb N$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.
Let $k=3^{2n+1}-2^{2n+1}-6^n$. By testing with numbers I got that $k$ cannot be prime for any value of $n$ (tested to $n=20$, so I assumed it's same for every value of $n$).
For $n=1$, $k<0$ $implies$ $k$ is not prime. (EDIT: oh... it's actually prime if $n=1$. Stupid me calculated $3^3=9$)
For $n>1$:
I managed to prove that if $n$ is even, $k$ is not a prime:
If $n$ is even, then $$3^{2n+1}equiv3 (mod 10),quad 2^{2n+1}equiv2 (mod 10),quad 6^nequiv6 (mod 10)$$
$$implies kequiv5 (mod 10)$$ So $k$ is divisible by 5 and therefore not prime.
Using the same method for if $n$ is odd, I get: $kequiv3 (mod 10)$, which doesn't prove anything.
How do I solve for when $n$ is odd?
elementary-number-theory prime-numbers modular-arithmetic
asked Dec 19 '18 at 20:16
PeroPero
1497
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marked as duplicate by Bill Dubuque modular-arithmetic
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Dec 19 '18 at 20:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44
add a comment |
$begingroup$
This question already has an answer here:
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
3 answers
Find all values of $n in mathbb N$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.
Let $k=3^{2n+1}-2^{2n+1}-6^n$. By testing with numbers I got that $k$ cannot be prime for any value of $n$ (tested to $n=20$, so I assumed it's same for every value of $n$).
For $n=1$, $k<0$ $implies$ $k$ is not prime. (EDIT: oh... it's actually prime if $n=1$. Stupid me calculated $3^3=9$)
For $n>1$:
I managed to prove that if $n$ is even, $k$ is not a prime:
If $n$ is even, then $$3^{2n+1}equiv3 (mod 10),quad 2^{2n+1}equiv2 (mod 10),quad 6^nequiv6 (mod 10)$$
$$implies kequiv5 (mod 10)$$ So $k$ is divisible by 5 and therefore not prime.
Using the same method for if $n$ is odd, I get: $kequiv3 (mod 10)$, which doesn't prove anything.
How do I solve for when $n$ is odd?
elementary-number-theory prime-numbers modular-arithmetic
asked Dec 19 '18 at 20:16
PeroPero
1497
$endgroup$
This question already has an answer here:
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
3 answers
Find all values of $n in mathbb N$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.
Let $k=3^{2n+1}-2^{2n+1}-6^n$. By testing with numbers I got that $k$ cannot be prime for any value of $n$ (tested to $n=20$, so I assumed it's same for every value of $n$).
For $n=1$, $k<0$ $implies$ $k$ is not prime. (EDIT: oh... it's actually prime if $n=1$. Stupid me calculated $3^3=9$)
For $n>1$:
I managed to prove that if $n$ is even, $k$ is not a prime:
If $n$ is even, then $$3^{2n+1}equiv3 (mod 10),quad 2^{2n+1}equiv2 (mod 10),quad 6^nequiv6 (mod 10)$$
$$implies kequiv5 (mod 10)$$ So $k$ is divisible by 5 and therefore not prime.
Using the same method for if $n$ is odd, I get: $kequiv3 (mod 10)$, which doesn't prove anything.
How do I solve for when $n$ is odd?
This question already has an answer here:
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
3 answers
elementary-number-theory prime-numbers modular-arithmetic
elementary-number-theory prime-numbers modular-arithmetic
asked Dec 19 '18 at 20:16
PeroPero
1497
asked Dec 19 '18 at 20:16
PeroPero
1497
edited Dec 19 '18 at 20:35
Pero
asked Dec 19 '18 at 20:16
PeroPero
1497
asked Dec 19 '18 at 20:16
PeroPero
1497
asked Dec 19 '18 at 20:16
PeroPero
1497
1497
marked as duplicate by Bill Dubuque modular-arithmetic
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Dec 19 '18 at 20:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque modular-arithmetic
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Dec 19 '18 at 20:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44
add a comment |
1
$begingroup$
Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44
1
1
$begingroup$
Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44
add a comment |
1 Answer
1
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$begingroup$
Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$
so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1implies (3-2)(3^{n-1}+3^{n-2}cdot 2+....+2^{n-1})=1$$
But this holds only if $n=1$.
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$
so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1implies (3-2)(3^{n-1}+3^{n-2}cdot 2+....+2^{n-1})=1$$
But this holds only if $n=1$.
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
add a comment |
$begingroup$
Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$
so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1implies (3-2)(3^{n-1}+3^{n-2}cdot 2+....+2^{n-1})=1$$
But this holds only if $n=1$.
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
add a comment |
$begingroup$
Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$
so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1implies (3-2)(3^{n-1}+3^{n-2}cdot 2+....+2^{n-1})=1$$
But this holds only if $n=1$.
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$
so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1implies (3-2)(3^{n-1}+3^{n-2}cdot 2+....+2^{n-1})=1$$
But this holds only if $n=1$.
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
edited Dec 19 '18 at 20:42
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
answered Dec 19 '18 at 20:22
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
add a comment |
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
1
1
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet.
$endgroup$
– fleablood
Dec 19 '18 at 20:30
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
$begingroup$
I explain how to derive this from scratch in this old answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:54
add a comment |
1
$begingroup$
Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$.
$endgroup$
– fleablood
Dec 19 '18 at 20:22
$begingroup$
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic).
$endgroup$
– Bill Dubuque
Dec 19 '18 at 20:41
$begingroup$
I'm seeing it now, thanks.
$endgroup$
– Pero
Dec 19 '18 at 20:44