Generating Prime Numbers From Composite Numbers
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1
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If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.
For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$
$ab-1=35times18-1=629$ which is a composite number, and
$ab+1=35times18+1=631$ which is a prime number.
Is the statement true?
elementary-number-theory prime-numbers
asked Nov 22 at 14:14
Hussain-Alqatari
2146
add a comment |
up vote
1
down vote
favorite
If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.
For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$
$ab-1=35times18-1=629$ which is a composite number, and
$ab+1=35times18+1=631$ which is a prime number.
Is the statement true?
elementary-number-theory prime-numbers
asked Nov 22 at 14:14
Hussain-Alqatari
2146
3
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
1
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.
For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$
$ab-1=35times18-1=629$ which is a composite number, and
$ab+1=35times18+1=631$ which is a prime number.
Is the statement true?
elementary-number-theory prime-numbers
asked Nov 22 at 14:14
Hussain-Alqatari
2146
If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.
For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$
$ab-1=35times18-1=629$ which is a composite number, and
$ab+1=35times18+1=631$ which is a prime number.
Is the statement true?
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
asked Nov 22 at 14:14
Hussain-Alqatari
2146
asked Nov 22 at 14:14
Hussain-Alqatari
2146
edited Nov 23 at 10:02
asked Nov 22 at 14:14
Hussain-Alqatari
2146
asked Nov 22 at 14:14
Hussain-Alqatari
2146
asked Nov 22 at 14:14
Hussain-Alqatari
2146
2146
3
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
1
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38
add a comment |
3
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
1
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38
3
3
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
1
1
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.
[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:
714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262
answered Nov 22 at 14:29
Especially Lime
21.1k22655
add a comment |
up vote
3
down vote
If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.
Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
$$
ab + 1 = 715 = 5cdot 11cdot 13\
ab - 1 = 713 = 23cdot 31
$$
answered Nov 22 at 14:19
Arthur
108k7103186
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.
[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:
714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262
answered Nov 22 at 14:29
Especially Lime
21.1k22655
add a comment |
up vote
2
down vote
accepted
You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.
[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:
714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262
answered Nov 22 at 14:29
Especially Lime
21.1k22655
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.
[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:
714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262
answered Nov 22 at 14:29
Especially Lime
21.1k22655
You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.
[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:
714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262
answered Nov 22 at 14:29
Especially Lime
21.1k22655
edited Nov 23 at 9:35
answered Nov 22 at 14:29
Especially Lime
21.1k22655
answered Nov 22 at 14:29
Especially Lime
21.1k22655
answered Nov 22 at 14:29
Especially Lime
21.1k22655
21.1k22655
add a comment |
add a comment |
up vote
3
down vote
If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.
Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
$$
ab + 1 = 715 = 5cdot 11cdot 13\
ab - 1 = 713 = 23cdot 31
$$
answered Nov 22 at 14:19
Arthur
108k7103186
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
add a comment |
up vote
3
down vote
If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.
Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
$$
ab + 1 = 715 = 5cdot 11cdot 13\
ab - 1 = 713 = 23cdot 31
$$
answered Nov 22 at 14:19
Arthur
108k7103186
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
add a comment |
up vote
3
down vote
up vote
3
down vote
If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.
Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
$$
ab + 1 = 715 = 5cdot 11cdot 13\
ab - 1 = 713 = 23cdot 31
$$
answered Nov 22 at 14:19
Arthur
108k7103186
If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.
Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
$$
ab + 1 = 715 = 5cdot 11cdot 13\
ab - 1 = 713 = 23cdot 31
$$
answered Nov 22 at 14:19
Arthur
108k7103186
edited Nov 22 at 14:29
answered Nov 22 at 14:19
Arthur
108k7103186
answered Nov 22 at 14:19
Arthur
108k7103186
answered Nov 22 at 14:19
Arthur
108k7103186
108k7103186
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
add a comment |
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
1
1
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
$a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
– Especially Lime
Nov 22 at 14:23
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
@EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
– Arthur
Nov 22 at 14:30
add a comment |
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3
$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16
1
If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38