Mixing
Is 31 the only number that can be represented by two distinct sums of consecutive powers of primes? [duplicate]
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On uniqueness of sums of prime powers
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I'm trying to prove that a number with two distinct prime factors can't be friends with another number with the same prime factors.
One way I could prove this is that there'd be only one example where $$sum_i=0^np^i=sum_j=0^mq^j$$
That example, preferrably, would be $2^0+2^1+2^2+2^3+2^4=5^0+5^1+5^2=31$, which fails to fit other conditions necessary to construct that pair of friends.
Through a little bit of computing power, I was unable to find examples for $p<300,n<10$, which leads me to believe it may be the only example. However, I'm completely lost on a continuation, if there is one, and whether this is just a case of the XY problem, and I should drop this line of reasoning and move elsewhere.
elementary-number-theory divisor-sum
asked Jul 15 at 6:51
Gelly
33
marked as duplicate by Steven Stadnicki, Trần Thúc Minh TrÃ, max_zorn, Claude Leibovici, Parcly Taxel Jul 17 at 11:35
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 |Â
up vote
0
down vote
favorite
This question already has an answer here:
On uniqueness of sums of prime powers
1 answer
I'm trying to prove that a number with two distinct prime factors can't be friends with another number with the same prime factors.
One way I could prove this is that there'd be only one example where $$sum_i=0^np^i=sum_j=0^mq^j$$
That example, preferrably, would be $2^0+2^1+2^2+2^3+2^4=5^0+5^1+5^2=31$, which fails to fit other conditions necessary to construct that pair of friends.
Through a little bit of computing power, I was unable to find examples for $p<300,n<10$, which leads me to believe it may be the only example. However, I'm completely lost on a continuation, if there is one, and whether this is just a case of the XY problem, and I should drop this line of reasoning and move elsewhere.
elementary-number-theory divisor-sum
asked Jul 15 at 6:51
Gelly
33
marked as duplicate by Steven Stadnicki, Trần Thúc Minh TrÃ, max_zorn, Claude Leibovici, Parcly Taxel Jul 17 at 11:35
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
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
On uniqueness of sums of prime powers
1 answer
I'm trying to prove that a number with two distinct prime factors can't be friends with another number with the same prime factors.
One way I could prove this is that there'd be only one example where $$sum_i=0^np^i=sum_j=0^mq^j$$
That example, preferrably, would be $2^0+2^1+2^2+2^3+2^4=5^0+5^1+5^2=31$, which fails to fit other conditions necessary to construct that pair of friends.
Through a little bit of computing power, I was unable to find examples for $p<300,n<10$, which leads me to believe it may be the only example. However, I'm completely lost on a continuation, if there is one, and whether this is just a case of the XY problem, and I should drop this line of reasoning and move elsewhere.
elementary-number-theory divisor-sum
asked Jul 15 at 6:51
Gelly
33
This question already has an answer here:
On uniqueness of sums of prime powers
1 answer
I'm trying to prove that a number with two distinct prime factors can't be friends with another number with the same prime factors.
One way I could prove this is that there'd be only one example where $$sum_i=0^np^i=sum_j=0^mq^j$$
That example, preferrably, would be $2^0+2^1+2^2+2^3+2^4=5^0+5^1+5^2=31$, which fails to fit other conditions necessary to construct that pair of friends.
Through a little bit of computing power, I was unable to find examples for $p<300,n<10$, which leads me to believe it may be the only example. However, I'm completely lost on a continuation, if there is one, and whether this is just a case of the XY problem, and I should drop this line of reasoning and move elsewhere.
This question already has an answer here:
On uniqueness of sums of prime powers
1 answer
elementary-number-theory divisor-sum
asked Jul 15 at 6:51
Gelly
33
asked Jul 15 at 6:51
Gelly
33
asked Jul 15 at 6:51
Gelly
33
asked Jul 15 at 6:51
Gelly
33
33
marked as duplicate by Steven Stadnicki, Trần Thúc Minh TrÃ, max_zorn, Claude Leibovici, Parcly Taxel Jul 17 at 11:35
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 Steven Stadnicki, Trần Thúc Minh TrÃ, max_zorn, Claude Leibovici, Parcly Taxel Jul 17 at 11:35
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
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05
add a comment |Â
1
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05
1
1
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Nope!
$$sum_i=0^290^i=1+90+90^2=8191=1+2+2^2+2^3+dots +2^12=sum_j=0^122^j$$
As is mentioned in the comments this question is duplicate.
answered Jul 15 at 7:14
Mason
1,2401224
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Nope!
$$sum_i=0^290^i=1+90+90^2=8191=1+2+2^2+2^3+dots +2^12=sum_j=0^122^j$$
As is mentioned in the comments this question is duplicate.
answered Jul 15 at 7:14
Mason
1,2401224
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
add a comment |Â
up vote
1
down vote
accepted
Nope!
$$sum_i=0^290^i=1+90+90^2=8191=1+2+2^2+2^3+dots +2^12=sum_j=0^122^j$$
As is mentioned in the comments this question is duplicate.
answered Jul 15 at 7:14
Mason
1,2401224
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Nope!
$$sum_i=0^290^i=1+90+90^2=8191=1+2+2^2+2^3+dots +2^12=sum_j=0^122^j$$
As is mentioned in the comments this question is duplicate.
answered Jul 15 at 7:14
Mason
1,2401224
Nope!
$$sum_i=0^290^i=1+90+90^2=8191=1+2+2^2+2^3+dots +2^12=sum_j=0^122^j$$
As is mentioned in the comments this question is duplicate.
answered Jul 15 at 7:14
Mason
1,2401224
answered Jul 15 at 7:14
Mason
1,2401224
answered Jul 15 at 7:14
Mason
1,2401224
answered Jul 15 at 7:14
Mason
1,2401224
1,2401224
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
add a comment |Â
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
Whoops! The OP asks for primes! This $90$ doesn't do the trick.
– Mason
Jul 16 at 21:16
add a comment |Â
1
Possibly related: math.stackexchange.com/questions/2599653/…
– Crostul
Jul 15 at 6:54
Possibly related?!?! You mean: Is the exact same question?
– Mason
Jul 15 at 7:02
Oh shoot, I did my best to see if this already had an answer. I guess I was looking with the wrong terms and being overdescriptive with it. The Goormaghtigh conjecture certainly helped in the other question, so thanks.
– Gelly
Jul 15 at 17:05