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Could a prime of the form $91^n+8264$ exist?
Do there exist any numbers such that any in-order combination of digits is prime (including the original number)?Prime factors of $sum_k=1^30k^k^k$Is there an obvious reason why $4^n+n^4$ cannot be prime for $nge 2$?Prime Number TestingDoes concatenating the perfect powers ever lead to a prime?Twin prime pairs of the form $(prod_j=1^n phi(j))pm 1$How to determine numbers in the form (6n+1) that are NOT prime?Can numbers of form $n^2 + 1$ have prime factors of form $4m + 3$?What is the largest known twin-prime of the form $2^acdot 3^bpm 1$?Is there a prime of the given form?
$begingroup$
Upto n=20.000, I did not find a prime of the form $91^n+8264$. Is there some mathematical reason why this type of prime cannot exist?
elementary-number-theory
edited 17 hours ago
Jyrki Lahtonen
110k13171387
asked yesterday
homunculushomunculus
1177
$endgroup$
add a comment |
$begingroup$
Upto n=20.000, I did not find a prime of the form $91^n+8264$. Is there some mathematical reason why this type of prime cannot exist?
elementary-number-theory
edited 17 hours ago
Jyrki Lahtonen
110k13171387
asked yesterday
homunculushomunculus
1177
$endgroup$
$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago
add a comment |
$begingroup$
Upto n=20.000, I did not find a prime of the form $91^n+8264$. Is there some mathematical reason why this type of prime cannot exist?
elementary-number-theory
edited 17 hours ago
Jyrki Lahtonen
110k13171387
asked yesterday
homunculushomunculus
1177
$endgroup$
Upto n=20.000, I did not find a prime of the form $91^n+8264$. Is there some mathematical reason why this type of prime cannot exist?
elementary-number-theory
elementary-number-theory
edited 17 hours ago
Jyrki Lahtonen
110k13171387
asked yesterday
homunculushomunculus
1177
edited 17 hours ago
Jyrki Lahtonen
110k13171387
asked yesterday
homunculushomunculus
1177
edited 17 hours ago
Jyrki Lahtonen
110k13171387
edited 17 hours ago
Jyrki Lahtonen
110k13171387
edited 17 hours ago
Jyrki Lahtonen
110k13171387
110k13171387
asked yesterday
homunculushomunculus
1177
asked yesterday
homunculushomunculus
1177
asked yesterday
homunculushomunculus
1177
1177
$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago
add a comment |
$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago
$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago
$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For any $ngeq 0$, $91^n$ will always end in a $1$, so $91^n + 8264$ will always end in a $5$.
answered yesterday
ArthurArthur
120k7120204
$endgroup$
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
add a comment |
$begingroup$
$$91^n+8264 = (3times 30+1)^n + (3times 2754+2)$$
Taking this (mod 3) gives 0 for all $nge 0$.
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
$endgroup$
add a comment |
$begingroup$
By below $91^n + 8264$ is divisible by $gcd(90,8265)=15,$ for all $,nge 0$
Theorem $ $ For all $, nge 0!: cmid a^n+b, iff, cmid a-1,b+1iff cmidgcd(a-1,b+1)$
Proof $ (Rightarrow) $ For $, n = 0,1,$ we have $,cmid 1+b,a+b,$ so $,c,$ divides their difference $, a-1.$
$(Leftarrow) a^n + b = a^n-1 + b+1,$ and $,cmid a-1mid a^n-1,$ for all $nge 0$
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any $ngeq 0$, $91^n$ will always end in a $1$, so $91^n + 8264$ will always end in a $5$.
answered yesterday
ArthurArthur
120k7120204
$endgroup$
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
add a comment |
$begingroup$
For any $ngeq 0$, $91^n$ will always end in a $1$, so $91^n + 8264$ will always end in a $5$.
answered yesterday
ArthurArthur
120k7120204
$endgroup$
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
add a comment |
$begingroup$
For any $ngeq 0$, $91^n$ will always end in a $1$, so $91^n + 8264$ will always end in a $5$.
answered yesterday
ArthurArthur
120k7120204
$endgroup$
For any $ngeq 0$, $91^n$ will always end in a $1$, so $91^n + 8264$ will always end in a $5$.
answered yesterday
ArthurArthur
120k7120204
edited 12 hours ago
answered yesterday
ArthurArthur
120k7120204
answered yesterday
ArthurArthur
120k7120204
answered yesterday
ArthurArthur
120k7120204
120k7120204
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
add a comment |
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
1
1
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
Actually, for $n geq 0$.
$endgroup$
– Alex Silva
13 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
$begingroup$
@AlexSilva You're right, although that's sortof just random happenstance.
$endgroup$
– Arthur
12 hours ago
add a comment |
$begingroup$
$$91^n+8264 = (3times 30+1)^n + (3times 2754+2)$$
Taking this (mod 3) gives 0 for all $nge 0$.
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
$endgroup$
add a comment |
$begingroup$
$$91^n+8264 = (3times 30+1)^n + (3times 2754+2)$$
Taking this (mod 3) gives 0 for all $nge 0$.
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
$endgroup$
add a comment |
$begingroup$
$$91^n+8264 = (3times 30+1)^n + (3times 2754+2)$$
Taking this (mod 3) gives 0 for all $nge 0$.
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
$endgroup$
$$91^n+8264 = (3times 30+1)^n + (3times 2754+2)$$
Taking this (mod 3) gives 0 for all $nge 0$.
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
answered yesterday
InterstellarProbeInterstellarProbe
3,049727
3,049727
add a comment |
add a comment |
$begingroup$
By below $91^n + 8264$ is divisible by $gcd(90,8265)=15,$ for all $,nge 0$
Theorem $ $ For all $, nge 0!: cmid a^n+b, iff, cmid a-1,b+1iff cmidgcd(a-1,b+1)$
Proof $ (Rightarrow) $ For $, n = 0,1,$ we have $,cmid 1+b,a+b,$ so $,c,$ divides their difference $, a-1.$
$(Leftarrow) a^n + b = a^n-1 + b+1,$ and $,cmid a-1mid a^n-1,$ for all $nge 0$
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
add a comment |
$begingroup$
By below $91^n + 8264$ is divisible by $gcd(90,8265)=15,$ for all $,nge 0$
Theorem $ $ For all $, nge 0!: cmid a^n+b, iff, cmid a-1,b+1iff cmidgcd(a-1,b+1)$
Proof $ (Rightarrow) $ For $, n = 0,1,$ we have $,cmid 1+b,a+b,$ so $,c,$ divides their difference $, a-1.$
$(Leftarrow) a^n + b = a^n-1 + b+1,$ and $,cmid a-1mid a^n-1,$ for all $nge 0$
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
add a comment |
$begingroup$
By below $91^n + 8264$ is divisible by $gcd(90,8265)=15,$ for all $,nge 0$
Theorem $ $ For all $, nge 0!: cmid a^n+b, iff, cmid a-1,b+1iff cmidgcd(a-1,b+1)$
Proof $ (Rightarrow) $ For $, n = 0,1,$ we have $,cmid 1+b,a+b,$ so $,c,$ divides their difference $, a-1.$
$(Leftarrow) a^n + b = a^n-1 + b+1,$ and $,cmid a-1mid a^n-1,$ for all $nge 0$
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
By below $91^n + 8264$ is divisible by $gcd(90,8265)=15,$ for all $,nge 0$
Theorem $ $ For all $, nge 0!: cmid a^n+b, iff, cmid a-1,b+1iff cmidgcd(a-1,b+1)$
Proof $ (Rightarrow) $ For $, n = 0,1,$ we have $,cmid 1+b,a+b,$ so $,c,$ divides their difference $, a-1.$
$(Leftarrow) a^n + b = a^n-1 + b+1,$ and $,cmid a-1mid a^n-1,$ for all $nge 0$
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
edited yesterday
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
answered yesterday
Bill DubuqueBill Dubuque
213k29195654
213k29195654
add a comment |
add a comment |
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$begingroup$
Did you check the factors of the numbers with $n=0,1,2,...$? Any factors repeating? If so, you are on your way. If not, then it may be more challenging :-)
$endgroup$
– Jyrki Lahtonen
17 hours ago