Prime sequence.
Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.
Best regards
prime-numbers modular-arithmetic
asked Dec 10 at 16:06
Dead_Ling0
406
add a comment |
Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.
Best regards
prime-numbers modular-arithmetic
asked Dec 10 at 16:06
Dead_Ling0
406
1
Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19
add a comment |
Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.
Best regards
prime-numbers modular-arithmetic
asked Dec 10 at 16:06
Dead_Ling0
406
Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.
Best regards
prime-numbers modular-arithmetic
prime-numbers modular-arithmetic
asked Dec 10 at 16:06
Dead_Ling0
406
asked Dec 10 at 16:06
Dead_Ling0
406
edited Dec 10 at 17:49
asked Dec 10 at 16:06
Dead_Ling0
406
asked Dec 10 at 16:06
Dead_Ling0
406
asked Dec 10 at 16:06
Dead_Ling0
406
406
1
Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19
add a comment |
1
Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19
1
1
Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19
add a comment |
4 Answers
4
active
oldest
votes
Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$
Therefore $p(p+2)+2 equiv 1 pmod {100}$
answered Dec 10 at 16:55
gammatester
16.7k21632
add a comment |
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
answered Dec 10 at 16:15
Arthur
110k7105186
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
add a comment |
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.
answered Dec 10 at 16:59
Keith Backman
9941612
add a comment |
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$
We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$
Therefore $p(p+2)+2 equiv 1 pmod {100}$
answered Dec 10 at 16:55
gammatester
16.7k21632
add a comment |
Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$
Therefore $p(p+2)+2 equiv 1 pmod {100}$
answered Dec 10 at 16:55
gammatester
16.7k21632
add a comment |
Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$
Therefore $p(p+2)+2 equiv 1 pmod {100}$
answered Dec 10 at 16:55
gammatester
16.7k21632
Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$
Therefore $p(p+2)+2 equiv 1 pmod {100}$
answered Dec 10 at 16:55
gammatester
16.7k21632
edited Dec 10 at 17:02
answered Dec 10 at 16:55
gammatester
16.7k21632
answered Dec 10 at 16:55
gammatester
16.7k21632
answered Dec 10 at 16:55
gammatester
16.7k21632
16.7k21632
add a comment |
add a comment |
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
answered Dec 10 at 16:15
Arthur
110k7105186
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
add a comment |
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
answered Dec 10 at 16:15
Arthur
110k7105186
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
add a comment |
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
answered Dec 10 at 16:15
Arthur
110k7105186
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
answered Dec 10 at 16:15
Arthur
110k7105186
edited Dec 10 at 16:25
answered Dec 10 at 16:15
Arthur
110k7105186
answered Dec 10 at 16:15
Arthur
110k7105186
answered Dec 10 at 16:15
Arthur
110k7105186
110k7105186
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
add a comment |
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
Will give that a look now thanks for posting :)
– Dead_Ling0
Dec 10 at 16:28
add a comment |
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.
answered Dec 10 at 16:59
Keith Backman
9941612
add a comment |
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.
answered Dec 10 at 16:59
Keith Backman
9941612
add a comment |
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.
answered Dec 10 at 16:59
Keith Backman
9941612
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.
answered Dec 10 at 16:59
Keith Backman
9941612
answered Dec 10 at 16:59
Keith Backman
9941612
answered Dec 10 at 16:59
Keith Backman
9941612
answered Dec 10 at 16:59
Keith Backman
9941612
9941612
add a comment |
add a comment |
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$
We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
add a comment |
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$
We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
add a comment |
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$
We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$
We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
edited Dec 10 at 20:16
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
answered Dec 10 at 16:12
Mark Bennet
80.5k981179
80.5k981179
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
add a comment |
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
How does mod 6 help any more than mod 2 with finding results mod 100?
– Arthur
Dec 10 at 16:25
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
@Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
– Mark Bennet
Dec 10 at 19:48
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
– Arthur
Dec 10 at 19:57
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
@Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
– Mark Bennet
Dec 10 at 20:09
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Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10
Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12
$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17
It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19