Fermat-Catalan conjecture - Revision history

Vipul: /* Other related conjectures and facts */

2010-08-13T19:04:04Z

Other related conjectures and facts

← Older revision		Revision as of 19:04, 13 August 2010
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	\| [[Fermat's last theorem]] \|\| <math>a^n + b^n = c^n</math> has no solutions for <math>n \ge 3</math> and <math>a,b,c</math> positive integers \|\| proved \|\| The Fermat-Catalan conjecture would imply that there are only finitely many solutions. The Fermat-Catalan conjecture along with an explicit list of solutions would prove Fermat's last theorem.		\| [[Fermat's last theorem]] \|\| <math>a^n + b^n = c^n</math> has no solutions for <math>n \ge 3</math> and <math>a,b,c</math> positive integers \|\| proved \|\| The Fermat-Catalan conjecture would imply that there are only finitely many solutions. The Fermat-Catalan conjecture along with an explicit list of solutions would prove Fermat's last theorem.
	\|-		\|-
	\| [[Beal conjecture]] \|\| <math>a^m + b^n = c^k</math> has no solutions for relatively prime <math>a,b,c</math> with <math>m,n,k</math> all at least <math>3</math> \|\| The Fermat-Catalan conjecture would imply that there are only finitely many solutions. If none of the solutions to the Fermat-Catalan problem provides a counterexample to the Beal conjecture, this would prove the Beal conjecture.		\| [[Beal conjecture]] \|\| <math>a^m + b^n = c^k</math> has no solutions for relatively prime <math>a,b,c</math> with <math>m,n,k</math> all at least <math>3</math> \|\| The Fermat-Catalan conjecture would imply that there are only finitely many solutions. If none of the solutions to the Fermat-Catalan problem provides a counterexample to the Beal conjecture, this would prove the Beal conjecture. \|\|
	\|-		\|-
	\| [[Faltings' theorem]] \|\| This states that certain kinds of algebraic curves have only finitely many rational points \|\| proved \|\| Faltings' theorem implies that for ''fixed'' choice of <math>m,n,k</math>, there are only finitely many solutions. The Fermat-Catalan conjecture goes further: it says that for all but finitely many permissible choices of <math>m,n,k</math>, there are ''no'' solutions.		\| [[Faltings' theorem]] \|\| This states that certain kinds of algebraic curves have only finitely many rational points \|\| proved \|\| Faltings' theorem implies that for ''fixed'' choice of <math>m,n,k</math>, there are only finitely many solutions. The Fermat-Catalan conjecture goes further: it says that for all but finitely many permissible choices of <math>m,n,k</math>, there are ''no'' solutions.
	\|}		\|}

Vipul: /* Status */

2010-08-13T19:03:37Z

Status

← Older revision		Revision as of 19:03, 13 August 2010
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	! Equation !! <math>a</math> !! <math>b</math> !! <math>c</math> !! <math>m</math> !! <math>n</math> !! <math>k</math> !! <math>a^m</math> !! <math>b^n</math> !! <math>c^k</math> !! Value of <math>1/m + 1/n + 1/k</math> !! Comment		! Equation !! <math>a</math> !! <math>b</math> !! <math>c</math> !! <math>m</math> !! <math>n</math> !! <math>k</math> !! <math>a^m</math> !! <math>b^n</math> !! <math>c^k</math> !! Value of <math>1/m + 1/n + 1/k</math> !! Comment
	\|-		\|-
	\| <math>\! 1^m + 2^3 = 3^2</math>, <math>m \ge 7</math> \|\| 1 \|\| 2 \|\| 3 \|\| ~~<math>m \ge 7</math>~~ \|\| 3 \|\| 2 \|\| 1 \|\| 8 \|\| 9 \|\| <math>(5/6) + (1/m)</math> \|\| [[Catalan's conjecture]] states that this is the only solution with <math>a = 1</math>. That conjecture has been proved.		\| <math>\! 1^m + 2^3 = 3^2</math>, <math>m \ge 7</math> \|\| 1 \|\| 2 \|\| 3 \|\| - \|\| 3 \|\| 2 \|\| 1 \|\| 8 \|\| 9 \|\| <math>(5/6) + (1/m)</math> \|\| [[Catalan's conjecture]] states that this is the only solution with <math>a = 1</math>. That conjecture has been proved.
	\|-		\|-
	\| <math>\! 2^5 + 7^2 = 3^4</math> \|\| 2 \|\| 7 \|\| 3 \|\| 5 \|\| 2 \|\| 4 \|\| 32 \|\| 49 \|\| 81 \|\| 0.95 = 19/20 \|\|		\| <math>\! 2^5 + 7^2 = 3^4</math> \|\| 2 \|\| 7 \|\| 3 \|\| 5 \|\| 2 \|\| 4 \|\| 32 \|\| 49 \|\| 81 \|\| 0.95 = 19/20 \|\|

Vipul: /* Status */

2010-08-13T19:03:07Z

Status

← Older revision		Revision as of 19:03, 13 August 2010
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	\| <math>\! 1^m + 2^3 = 3^2</math>, <math>m \ge 7</math> \|\| 1 \|\| 2 \|\| 3 \|\| <math>m \ge 7</math> \|\| 3 \|\| 2 \|\| 1 \|\| 8 \|\| 9 \|\| <math>(5/6) + (1/m)</math> \|\| [[Catalan's conjecture]] states that this is the only solution with <math>a = 1</math>. That conjecture has been proved.		\| <math>\! 1^m + 2^3 = 3^2</math>, <math>m \ge 7</math> \|\| 1 \|\| 2 \|\| 3 \|\| <math>m \ge 7</math> \|\| 3 \|\| 2 \|\| 1 \|\| 8 \|\| 9 \|\| <math>(5/6) + (1/m)</math> \|\| [[Catalan's conjecture]] states that this is the only solution with <math>a = 1</math>. That conjecture has been proved.
	\|-		\|-
	\| <math>\! 2^5 + 7^2 = 3^4</math> \|\| 2 \|\| 7 \|\| 3 \|\| 5 \|\| 2 \|\| 4 \|\| 32 \|\| 49 \|\| 81 \|\| 0.95 = 19/20		\| <math>\! 2^5 + 7^2 = 3^4</math> \|\| 2 \|\| 7 \|\| 3 \|\| 5 \|\| 2 \|\| 4 \|\| 32 \|\| 49 \|\| 81 \|\| 0.95 = 19/20 \|\|
	\|-		\|-
	\| <math>\! 13^2 + 7^3 = 2^9</math> \|\| 13 \|\| 7 \|\| 2 \|\| 2 \|\| 3 \|\| 9 \|\| 169 \|\| 343 \|\| 512 \|\| 17/18		\| <math>\! 13^2 + 7^3 = 2^9</math> \|\| 13 \|\| 7 \|\| 2 \|\| 2 \|\| 3 \|\| 9 \|\| 169 \|\| 343 \|\| 512 \|\| 17/18 \|\|
	\|-		\|-
	\| <math>\! 2^7 + 17^3 = 71^2</math> \|\| 2 \|\| 17 \|\| 71 \|\| 7 \|\| 3 \|\| 2 \|\| 128 \|\| 4913 \|\| 5041 \|\| 41/42		\| <math>\! 2^7 + 17^3 = 71^2</math> \|\| 2 \|\| 17 \|\| 71 \|\| 7 \|\| 3 \|\| 2 \|\| 128 \|\| 4913 \|\| 5041 \|\| 41/42\|\|
	\|-		\|-
	\| <math>\! 3^5 + 11^4 = 122^2</math> \|\| 3 \|\| 11 \|\| 122 \|\| 5 \|\| 4 \|\| 2 \|\| 243 \|\| 14641 \|\| 14884 \|\| 0.95 = 19/20		\| <math>\! 3^5 + 11^4 = 122^2</math> \|\| 3 \|\| 11 \|\| 122 \|\| 5 \|\| 4 \|\| 2 \|\| 243 \|\| 14641 \|\| 14884 \|\| 0.95 = 19/20 \|\|
	\|-		\|-
	\| <math>\! 33^8 + 1549034^2 = 15613^3</math> \|\| 33 \|\| 1549034 \|\| 15613 \|\| 8 \|\| 2 \|\| 3 \|\| 1406408618241 \|\| 2399506333156 \|\| 3805914951397 \|\| 23/24		\| <math>\! 33^8 + 1549034^2 = 15613^3</math> \|\| 33 \|\| 1549034 \|\| 15613 \|\| 8 \|\| 2 \|\| 3 \|\| 1406408618241 \|\| 2399506333156 \|\| 3805914951397 \|\| 23/24 \|\|
	\|-		\|-
	\| <math>\! 1414^3 + 2213459^2 = 65^7</math> \|\| 1414 \|\| 2213459 \|\| 65 \|\| 3 \|\| 2 \|\| 7 \|\| 2827145944 \|\| 4899400744681 \|\| 4902227890625 \|\| 41/42		\| <math>\! 1414^3 + 2213459^2 = 65^7</math> \|\| 1414 \|\| 2213459 \|\| 65 \|\| 3 \|\| 2 \|\| 7 \|\| 2827145944 \|\| 4899400744681 \|\| 4902227890625 \|\| 41/42 \|\|
	\|-		\|-
	\| <math>\! 9262^3 + 15312283^2 = 113^7</math> \|\| 9262 \|\| 15312283 \|\| 113 \|\| 3 \|\| 2 \|\| 7 \|\| 794537372728 \|\| 234466010672089 \|\| 235260548044817 \|\| 41/42		\| <math>\! 9262^3 + 15312283^2 = 113^7</math> \|\| 9262 \|\| 15312283 \|\| 113 \|\| 3 \|\| 2 \|\| 7 \|\| 794537372728 \|\| 234466010672089 \|\| 235260548044817 \|\| 41/42 \|\|
	\|-		\|-
	\| <math>\! 17^7 + 76271^3 = 21063928^2</math> \|\| 17 \|\| 76271 \|\| 21063928 \|\| 7 \|\| 3 \|\| 2 \|\| 410338673 \|\| 443688652450511 \|\| 443689062789184 \|\| 41/42		\| <math>\! 17^7 + 76271^3 = 21063928^2</math> \|\| 17 \|\| 76271 \|\| 21063928 \|\| 7 \|\| 3 \|\| 2 \|\| 410338673 \|\| 443688652450511 \|\| 443689062789184 \|\| 41/42 \|\|
	\|-		\|-
	\| <math>\! 43^8 + 96222^3 = 30042907^2</math> \|\| 43 \|\| 96222 \|\| 30042907 \|\| 8 \|\| 3 \|\| 2 \|\| 11688200277601 \|\| 890888060733048 \|\| 902576261010649 \|\| 23/24		\| <math>\! 43^8 + 96222^3 = 30042907^2</math> \|\| 43 \|\| 96222 \|\| 30042907 \|\| 8 \|\| 3 \|\| 2 \|\| 11688200277601 \|\| 890888060733048 \|\| 902576261010649 \|\| 23/24 \|\|
	\|}		\|}

Vipul: Created page with "==Statement== The equation: $\! a^m + b^n = c^k$ with $a,b,c,m,n,k$ positive integers such that $a,b,c$ are relatively prime and: $\frac..."$

2010-08-13T17:52:34Z

Created page with "==Statement== The equation: <math>\! a^m + b^n = c^k</math> with <math>a,b,c,m,n,k</math> positive integers such that <math>a,b,c</math> are relatively prime and: <math>\frac..."

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==Statement==

The equation:

<math>\! a^m + b^n = c^k</math>

with <math>a,b,c,m,n,k</math> positive integers such that <math>a,b,c</math> are relatively prime and:

<math>\frac{1}{m} + \frac{1}{n} + \frac{1}{k} \le 1</math>

has only finitely many solutions, where, in the special case that <math>a = 1</math> or <math>b = 1</math>, we count all solutions with distinct values of the corresponding exponent as equivalent.

(Note that sometimes the conjecture is written with strict inequality <math>< 1</math>. This is an equivalent formulation because in all the cases of exact equality with <math>1</math>, it is known that there are no solutions).

==Status==

The conjecture is open. A certain finite list of solutions has been found, but the conjecture does not claim that this is the complete list. Note that to avoid unnecessary duplication, we do not list the mirror solutions to each solution thatcan be obtained by interchanging <math>a</math> with <math>b</math> and <math>m</math> with <math>n</math>. The solutions found so far are:

{| class="sortable" border="1"
! Equation !! <math>a</math> !! <math>b</math> !! <math>c</math> !! <math>m</math> !! <math>n</math> !! <math>k</math> !! <math>a^m</math> !! <math>b^n</math> !! <math>c^k</math> !! Value of <math>1/m + 1/n + 1/k</math> !! Comment
|-
| <math>\! 1^m + 2^3 = 3^2</math>, <math>m \ge 7</math> || 1 || 2 || 3 || <math>m \ge 7</math> || 3 || 2 || 1 || 8 || 9 || <math>(5/6) + (1/m)</math> || [[Catalan's conjecture]] states that this is the only solution with <math>a = 1</math>. That conjecture has been proved.
|-
| <math>\! 2^5 + 7^2 = 3^4</math> || 2 || 7 || 3 || 5 || 2 || 4 || 32 || 49 || 81 || 0.95 = 19/20
|-
| <math>\! 13^2 + 7^3 = 2^9</math> || 13 || 7 || 2 || 2 || 3 || 9 || 169 || 343 || 512 || 17/18
|-
| <math>\! 2^7 + 17^3 = 71^2</math> || 2 || 17 || 71 || 7 || 3 || 2 || 128 || 4913 || 5041 || 41/42
|-
| <math>\! 3^5 + 11^4 = 122^2</math> || 3 || 11 || 122 || 5 || 4 || 2 || 243 || 14641 || 14884 || 0.95 = 19/20
|-
| <math>\! 33^8 + 1549034^2 = 15613^3</math> || 33 || 1549034 || 15613 || 8 || 2 || 3 || 1406408618241 || 2399506333156 || 3805914951397 || 23/24
|-
| <math>\! 1414^3 + 2213459^2 = 65^7</math> || 1414 || 2213459 || 65 || 3 || 2 || 7 || 2827145944 || 4899400744681 || 4902227890625 || 41/42
|-
| <math>\! 9262^3 + 15312283^2 = 113^7</math> || 9262 || 15312283 || 113 || 3 || 2 || 7 || 794537372728 || 234466010672089 || 235260548044817 || 41/42
|-
| <math>\! 17^7 + 76271^3 = 21063928^2</math> || 17 || 76271 || 21063928 || 7 || 3 || 2 || 410338673 || 443688652450511 || 443689062789184 || 41/42
|-
| <math>\! 43^8 + 96222^3 = 30042907^2</math> || 43 || 96222 || 30042907 || 8 || 3 || 2 || 11688200277601 || 890888060733048 || 902576261010649 || 23/24
|}

==Relation with other facts/conjectures==

===Stronger facts and conjectures===

{| class="sortable" border="1"
! Conjecture !! Statement !! Status !! Proof of implication
|-
| [[Weaker than::abc conjecture]] || For every <math>\epsilon > 0</math>,there exists <math>C_\epsilon</math> such that for all <math>a + b = c</math>, <math>\max \{ |a|, |b|, |c| \} \le C_\epsilon \prod_{p|abc}p^{1 + \epsilon}</math> || open || [[abc conjecture implies Fermat-Catalan conjecture]]
|}

===Other related conjectures and facts===

{| class="sortable" border="1"
! Conjecture/fact !! Statement !! Status !! Nature of relationship
|-
| [[Catalan's conjecture]] || The only solution to <math>1 + a^m = b^n</math> for <math>m,n > 1</math> and <math>a,b</math> positive is <math>1 + 2^3 = 3^2</math> || proved || If a complete set of solutions to the Fermat-Catalan conjecture is found and proved to be complete, Catalan's conjecture would be a special case.
|-
| [[Fermat's last theorem]] || <math>a^n + b^n = c^n</math> has no solutions for <math>n \ge 3</math> and <math>a,b,c</math> positive integers || proved || The Fermat-Catalan conjecture would imply that there are only finitely many solutions. The Fermat-Catalan conjecture along with an explicit list of solutions would prove Fermat's last theorem.
|-
| [[Beal conjecture]] || <math>a^m + b^n = c^k</math> has no solutions for relatively prime <math>a,b,c</math> with <math>m,n,k</math> all at least <math>3</math> || The Fermat-Catalan conjecture would imply that there are only finitely many solutions. If none of the solutions to the Fermat-Catalan problem provides a counterexample to the Beal conjecture, this would prove the Beal conjecture.
|-
| [[Faltings' theorem]] || This states that certain kinds of algebraic curves have only finitely many rational points || proved || Faltings' theorem implies that for ''fixed'' choice of <math>m,n,k</math>, there are only finitely many solutions. The Fermat-Catalan conjecture goes further: it says that for all but finitely many permissible choices of <math>m,n,k</math>, there are ''no'' solutions.
|}

Fermat-Catalan conjecture - Revision history

Vipul: /* Other related conjectures and facts */

Vipul: /* Status */

Vipul: /* Status */

Vipul: Created page with "==Statement== The equation: \! a^m + b^n = c^k with a,b,c,m,n,k positive integers such that a,b,c are relatively prime and: \frac..."

Vipul: Created page with "==Statement== The equation: $\! a^m + b^n = c^k$ with $a,b,c,m,n,k$ positive integers such that $a,b,c$ are relatively prime and: $\frac..."$