On uniqueness of sums of prime powers
$begingroup$
An exercise in number theory led to me to the following problem:
Find all solutions $(p,n,q,m)$ of the following equation:
$$sum_{k=0}^n p^k = sum_{h=0}^m q^h,$$
where $p<q$ are distinct primes, and $1 le m < n$ are indeterminates.
Numerical evidence gives me the only solution $$(p,n,q,m)=(2,4,5,2).$$
There might be no other solution: I have no idea on how to show it. For those interested of the source of this equation, here it comes the interesting exercise in number theory.
Find all numbers $A$ such that the sum of divisors of $A$ divisible by
$5$ equals the sum of divisors of $A$ divisible by $2$: $$sum_{2|d|A}
> d = sum_{5|e|A} e. $$
Clearly this is equivalent to the condition that the sum of divisors of $A$ NOT divisible by $5$ equals the sum of divisors of $A$ NOT divisible by $2$.
Let's factorize $A= 2^n cdot 5^m cdot w$, with $n,m ge 0$ and $gcd(w,10)=1$. Without loss fo generality we can consider $n,mneq 0$. Then we get
$$sum_{d|2^nw} d = sum_{e|5^mw} e$$
which is equivalent (using multiplicativity of the sum-of-divisors-function) to
$$sum_{k=0}^n 2^k = sum_{h=0}^m 5^h$$
Giving us $A=2^4 cdot 5^2 cdot w=400w$ (with $gcd(w,10)=1$).
Thus, I am looking for simple generalizations with arbirary primes $p neq q$.
number-theory prime-numbers
edited Jun 11 '18 at 8:01
Klangen
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
$endgroup$
add a comment |
$begingroup$
An exercise in number theory led to me to the following problem:
Find all solutions $(p,n,q,m)$ of the following equation:
$$sum_{k=0}^n p^k = sum_{h=0}^m q^h,$$
where $p<q$ are distinct primes, and $1 le m < n$ are indeterminates.
Numerical evidence gives me the only solution $$(p,n,q,m)=(2,4,5,2).$$
There might be no other solution: I have no idea on how to show it. For those interested of the source of this equation, here it comes the interesting exercise in number theory.
Find all numbers $A$ such that the sum of divisors of $A$ divisible by
$5$ equals the sum of divisors of $A$ divisible by $2$: $$sum_{2|d|A}
> d = sum_{5|e|A} e. $$
Clearly this is equivalent to the condition that the sum of divisors of $A$ NOT divisible by $5$ equals the sum of divisors of $A$ NOT divisible by $2$.
Let's factorize $A= 2^n cdot 5^m cdot w$, with $n,m ge 0$ and $gcd(w,10)=1$. Without loss fo generality we can consider $n,mneq 0$. Then we get
$$sum_{d|2^nw} d = sum_{e|5^mw} e$$
which is equivalent (using multiplicativity of the sum-of-divisors-function) to
$$sum_{k=0}^n 2^k = sum_{h=0}^m 5^h$$
Giving us $A=2^4 cdot 5^2 cdot w=400w$ (with $gcd(w,10)=1$).
Thus, I am looking for simple generalizations with arbirary primes $p neq q$.
number-theory prime-numbers
edited Jun 11 '18 at 8:01
Klangen
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
$endgroup$
2
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
7
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18
add a comment |
$begingroup$
An exercise in number theory led to me to the following problem:
Find all solutions $(p,n,q,m)$ of the following equation:
$$sum_{k=0}^n p^k = sum_{h=0}^m q^h,$$
where $p<q$ are distinct primes, and $1 le m < n$ are indeterminates.
Numerical evidence gives me the only solution $$(p,n,q,m)=(2,4,5,2).$$
There might be no other solution: I have no idea on how to show it. For those interested of the source of this equation, here it comes the interesting exercise in number theory.
Find all numbers $A$ such that the sum of divisors of $A$ divisible by
$5$ equals the sum of divisors of $A$ divisible by $2$: $$sum_{2|d|A}
> d = sum_{5|e|A} e. $$
Clearly this is equivalent to the condition that the sum of divisors of $A$ NOT divisible by $5$ equals the sum of divisors of $A$ NOT divisible by $2$.
Let's factorize $A= 2^n cdot 5^m cdot w$, with $n,m ge 0$ and $gcd(w,10)=1$. Without loss fo generality we can consider $n,mneq 0$. Then we get
$$sum_{d|2^nw} d = sum_{e|5^mw} e$$
which is equivalent (using multiplicativity of the sum-of-divisors-function) to
$$sum_{k=0}^n 2^k = sum_{h=0}^m 5^h$$
Giving us $A=2^4 cdot 5^2 cdot w=400w$ (with $gcd(w,10)=1$).
Thus, I am looking for simple generalizations with arbirary primes $p neq q$.
number-theory prime-numbers
edited Jun 11 '18 at 8:01
Klangen
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
$endgroup$
An exercise in number theory led to me to the following problem:
Find all solutions $(p,n,q,m)$ of the following equation:
$$sum_{k=0}^n p^k = sum_{h=0}^m q^h,$$
where $p<q$ are distinct primes, and $1 le m < n$ are indeterminates.
Numerical evidence gives me the only solution $$(p,n,q,m)=(2,4,5,2).$$
There might be no other solution: I have no idea on how to show it. For those interested of the source of this equation, here it comes the interesting exercise in number theory.
Find all numbers $A$ such that the sum of divisors of $A$ divisible by
$5$ equals the sum of divisors of $A$ divisible by $2$: $$sum_{2|d|A}
> d = sum_{5|e|A} e. $$
Clearly this is equivalent to the condition that the sum of divisors of $A$ NOT divisible by $5$ equals the sum of divisors of $A$ NOT divisible by $2$.
Let's factorize $A= 2^n cdot 5^m cdot w$, with $n,m ge 0$ and $gcd(w,10)=1$. Without loss fo generality we can consider $n,mneq 0$. Then we get
$$sum_{d|2^nw} d = sum_{e|5^mw} e$$
which is equivalent (using multiplicativity of the sum-of-divisors-function) to
$$sum_{k=0}^n 2^k = sum_{h=0}^m 5^h$$
Giving us $A=2^4 cdot 5^2 cdot w=400w$ (with $gcd(w,10)=1$).
Thus, I am looking for simple generalizations with arbirary primes $p neq q$.
number-theory prime-numbers
number-theory prime-numbers
edited Jun 11 '18 at 8:01
Klangen
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
edited Jun 11 '18 at 8:01
Klangen
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
edited Jun 11 '18 at 8:01
Klangen
1,73711334
edited Jun 11 '18 at 8:01
Klangen
1,73711334
edited Jun 11 '18 at 8:01
Klangen
1,73711334
1,73711334
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
asked Jan 10 '18 at 14:20
CrostulCrostul
28.1k22352
28.1k22352
2
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
7
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18
add a comment |
2
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
7
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18
2
2
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
7
7
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Promoting my comment to an answer: the question here is almost exactly the Goormaghtigh Conjecture — the only difference is that the Goormaghtigh Conjecture is just slightly more general and drops the primality condition on the two bases. Unfortunately, none of the references I'm able to find for it suggests that the special case is any simpler than the general conjecture (it's not at all clear how to make use of the primality of the bases to simplify the problem), and given the sparsity of the results, it seems as though the conjecture isn't attracting too much active research.
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
$endgroup$
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2599653%2fon-uniqueness-of-sums-of-prime-powers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Promoting my comment to an answer: the question here is almost exactly the Goormaghtigh Conjecture — the only difference is that the Goormaghtigh Conjecture is just slightly more general and drops the primality condition on the two bases. Unfortunately, none of the references I'm able to find for it suggests that the special case is any simpler than the general conjecture (it's not at all clear how to make use of the primality of the bases to simplify the problem), and given the sparsity of the results, it seems as though the conjecture isn't attracting too much active research.
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
$endgroup$
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
add a comment |
$begingroup$
Promoting my comment to an answer: the question here is almost exactly the Goormaghtigh Conjecture — the only difference is that the Goormaghtigh Conjecture is just slightly more general and drops the primality condition on the two bases. Unfortunately, none of the references I'm able to find for it suggests that the special case is any simpler than the general conjecture (it's not at all clear how to make use of the primality of the bases to simplify the problem), and given the sparsity of the results, it seems as though the conjecture isn't attracting too much active research.
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
$endgroup$
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
add a comment |
$begingroup$
Promoting my comment to an answer: the question here is almost exactly the Goormaghtigh Conjecture — the only difference is that the Goormaghtigh Conjecture is just slightly more general and drops the primality condition on the two bases. Unfortunately, none of the references I'm able to find for it suggests that the special case is any simpler than the general conjecture (it's not at all clear how to make use of the primality of the bases to simplify the problem), and given the sparsity of the results, it seems as though the conjecture isn't attracting too much active research.
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
$endgroup$
Promoting my comment to an answer: the question here is almost exactly the Goormaghtigh Conjecture — the only difference is that the Goormaghtigh Conjecture is just slightly more general and drops the primality condition on the two bases. Unfortunately, none of the references I'm able to find for it suggests that the special case is any simpler than the general conjecture (it's not at all clear how to make use of the primality of the bases to simplify the problem), and given the sparsity of the results, it seems as though the conjecture isn't attracting too much active research.
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
answered Jul 16 '18 at 17:17
Steven StadnickiSteven Stadnicki
41.3k868122
41.3k868122
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
add a comment |
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
1
1
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
$begingroup$
@mason That particular solution is not a solution to OP's specific question, though, since $90$ isn't prime.
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 21:11
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2599653%2fon-uniqueness-of-sums-of-prime-powers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
$$sum_{k=0}^n p^k = frac{p^{n+1} - 1}{p-1}$$
$endgroup$
– orlp
Jan 10 '18 at 15:46
7
$begingroup$
What you've got is (almost) exactly en.wikipedia.org/wiki/Goormaghtigh_conjecture (if you're curious, I found this because I remembered something recently on MO about the Feit-Thompson conjecture and its relation to the lack of odd finite simple groups, did a little bit of digging on the latter, and the Wikipedia page for that conjecture points to this one.)
$endgroup$
– Steven Stadnicki
Jan 10 '18 at 18:11
$begingroup$
@StevenStadnicki. Your comment should be promoted to answer. And then we should delete this dupe: math.stackexchange.com/questions/2852249/…
$endgroup$
– Mason
Jul 15 '18 at 22:34
$begingroup$
@Mason Done! (And I will vote on dup closure as soon as that's feasible)
$endgroup$
– Steven Stadnicki
Jul 16 '18 at 17:18