The sum of two prime powers equal a third prime power.












-1












$begingroup$


Is



$$13^2 + 7^3 = 512=2^9$$



the only solution for the sum of two primes $p,q$ raised to powers greater than $1$ equals a third prime power?






prime-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, there is $2^2+2^2=2^3$.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 21:53






  • 2




    $begingroup$
    begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
    $endgroup$
    – Donald Splutterwit
    Dec 16 '18 at 21:55






  • 1




    $begingroup$
    Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 21:57
















-1












$begingroup$


Is



$$13^2 + 7^3 = 512=2^9$$



the only solution for the sum of two primes $p,q$ raised to powers greater than $1$ equals a third prime power?






prime-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, there is $2^2+2^2=2^3$.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 21:53






  • 2




    $begingroup$
    begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
    $endgroup$
    – Donald Splutterwit
    Dec 16 '18 at 21:55






  • 1




    $begingroup$
    Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 21:57














-1












-1








-1





$begingroup$


Is



$$13^2 + 7^3 = 512=2^9$$



the only solution for the sum of two primes $p,q$ raised to powers greater than $1$ equals a third prime power?






prime-numbers







share|cite|improve this question











$endgroup$




Is



$$13^2 + 7^3 = 512=2^9$$



the only solution for the sum of two primes $p,q$ raised to powers greater than $1$ equals a third prime power?






prime-numbers




prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 16:57









Klangen

1,74411334




1,74411334










asked Dec 16 '18 at 21:43









J. M. BergotJ. M. Bergot

40528




40528












  • $begingroup$
    Well, there is $2^2+2^2=2^3$.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 21:53






  • 2




    $begingroup$
    begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
    $endgroup$
    – Donald Splutterwit
    Dec 16 '18 at 21:55






  • 1




    $begingroup$
    Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 21:57


















  • $begingroup$
    Well, there is $2^2+2^2=2^3$.
    $endgroup$
    – Oscar Lanzi
    Dec 16 '18 at 21:53






  • 2




    $begingroup$
    begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
    $endgroup$
    – Donald Splutterwit
    Dec 16 '18 at 21:55






  • 1




    $begingroup$
    Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 21:57
















$begingroup$
Well, there is $2^2+2^2=2^3$.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 21:53




$begingroup$
Well, there is $2^2+2^2=2^3$.
$endgroup$
– Oscar Lanzi
Dec 16 '18 at 21:53




2




2




$begingroup$
begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
$endgroup$
– Donald Splutterwit
Dec 16 '18 at 21:55




$begingroup$
begin{eqnarray*} 2^5+7^2=3^4 end{eqnarray*} There are other nice examples here ... people.math.sfu.ca/~ichen/pub/BeCh2.pdf
$endgroup$
– Donald Splutterwit
Dec 16 '18 at 21:55




1




1




$begingroup$
Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
$endgroup$
– Lucas Henrique
Dec 16 '18 at 21:57




$begingroup$
Since $1+1 not equiv 1, rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b pm q^c$.
$endgroup$
– Lucas Henrique
Dec 16 '18 at 21:57










2 Answers
2






active

oldest

votes


















1












$begingroup$

No, it isn't. For instance, we have:





  • $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and


  • $2^4+3^2=5^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 16 '18 at 22:16






  • 1




    $begingroup$
    The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
    $endgroup$
    – J. M. Bergot
    Dec 18 '18 at 18:43



















0












$begingroup$

You can check these types of numbers on wolfram alpha.



However you can see that $3^1+6^1=9^1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 19 '18 at 20:06











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

No, it isn't. For instance, we have:





  • $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and


  • $2^4+3^2=5^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 16 '18 at 22:16






  • 1




    $begingroup$
    The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
    $endgroup$
    – J. M. Bergot
    Dec 18 '18 at 18:43
















1












$begingroup$

No, it isn't. For instance, we have:





  • $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and


  • $2^4+3^2=5^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 16 '18 at 22:16






  • 1




    $begingroup$
    The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
    $endgroup$
    – J. M. Bergot
    Dec 18 '18 at 18:43














1












1








1





$begingroup$

No, it isn't. For instance, we have:





  • $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and


  • $2^4+3^2=5^2$.






share|cite|improve this answer











$endgroup$



No, it isn't. For instance, we have:





  • $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and


  • $2^4+3^2=5^2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 20:06

























answered Dec 16 '18 at 21:57









Oscar LanziOscar Lanzi

13.2k12136




13.2k12136












  • $begingroup$
    Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 16 '18 at 22:16






  • 1




    $begingroup$
    The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
    $endgroup$
    – J. M. Bergot
    Dec 18 '18 at 18:43


















  • $begingroup$
    Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 16 '18 at 22:16






  • 1




    $begingroup$
    The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
    $endgroup$
    – J. M. Bergot
    Dec 18 '18 at 18:43
















$begingroup$
Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
$endgroup$
– J. M. Bergot
Dec 16 '18 at 22:16




$begingroup$
Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2.
$endgroup$
– J. M. Bergot
Dec 16 '18 at 22:16




1




1




$begingroup$
The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
$endgroup$
– J. M. Bergot
Dec 18 '18 at 18:43




$begingroup$
The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity.
$endgroup$
– J. M. Bergot
Dec 18 '18 at 18:43











0












$begingroup$

You can check these types of numbers on wolfram alpha.



However you can see that $3^1+6^1=9^1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 19 '18 at 20:06
















0












$begingroup$

You can check these types of numbers on wolfram alpha.



However you can see that $3^1+6^1=9^1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 19 '18 at 20:06














0












0








0





$begingroup$

You can check these types of numbers on wolfram alpha.



However you can see that $3^1+6^1=9^1$






share|cite|improve this answer











$endgroup$



You can check these types of numbers on wolfram alpha.



However you can see that $3^1+6^1=9^1$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 19:21









Tianlalu

3,08421138




3,08421138










answered Dec 18 '18 at 19:09







user627661



















  • $begingroup$
    It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 19 '18 at 20:06


















  • $begingroup$
    It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
    $endgroup$
    – J. M. Bergot
    Dec 19 '18 at 20:06
















$begingroup$
It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
$endgroup$
– J. M. Bergot
Dec 19 '18 at 20:06




$begingroup$
It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2.
$endgroup$
– J. M. Bergot
Dec 19 '18 at 20:06


















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