Find three prime numbers












1












$begingroup$


I have to find three prime numbers where, the product of them is equal to seven times of sum of them.



So I wrote the equation:



$x$-1st of them;



$y$-2nd;



$z$-3rd;



$xyz=7(x+y+z)$



${xyzover 7}=x+y+z$



Where $x,y,z$ are prime numbers, so one of them must be $7$.



For example $x$



$yz=(7+y+z)$



Now I have to find these two numbers



I have found $5$ and $3$



$3*5*7=7(15)$



Which is correct.



My question is; Are there any other prime numbers, which
meet the equation?






prime-numbers systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:21












  • $begingroup$
    Yeah, I wrote it, but a little differently ;)
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:23










  • $begingroup$
    I don't see where you write that we are done, not even a little differently:)
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:24












  • $begingroup$
    If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:26








  • 2




    $begingroup$
    $yz=y+z+7$ is the same as $(y-1)(z-1)=8$
    $endgroup$
    – Empy2
    Dec 2 '18 at 12:36
















1












$begingroup$


I have to find three prime numbers where, the product of them is equal to seven times of sum of them.



So I wrote the equation:



$x$-1st of them;



$y$-2nd;



$z$-3rd;



$xyz=7(x+y+z)$



${xyzover 7}=x+y+z$



Where $x,y,z$ are prime numbers, so one of them must be $7$.



For example $x$



$yz=(7+y+z)$



Now I have to find these two numbers



I have found $5$ and $3$



$3*5*7=7(15)$



Which is correct.



My question is; Are there any other prime numbers, which
meet the equation?






prime-numbers systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:21












  • $begingroup$
    Yeah, I wrote it, but a little differently ;)
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:23










  • $begingroup$
    I don't see where you write that we are done, not even a little differently:)
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:24












  • $begingroup$
    If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:26








  • 2




    $begingroup$
    $yz=y+z+7$ is the same as $(y-1)(z-1)=8$
    $endgroup$
    – Empy2
    Dec 2 '18 at 12:36














1












1








1





$begingroup$


I have to find three prime numbers where, the product of them is equal to seven times of sum of them.



So I wrote the equation:



$x$-1st of them;



$y$-2nd;



$z$-3rd;



$xyz=7(x+y+z)$



${xyzover 7}=x+y+z$



Where $x,y,z$ are prime numbers, so one of them must be $7$.



For example $x$



$yz=(7+y+z)$



Now I have to find these two numbers



I have found $5$ and $3$



$3*5*7=7(15)$



Which is correct.



My question is; Are there any other prime numbers, which
meet the equation?






prime-numbers systems-of-equations







share|cite|improve this question











$endgroup$




I have to find three prime numbers where, the product of them is equal to seven times of sum of them.



So I wrote the equation:



$x$-1st of them;



$y$-2nd;



$z$-3rd;



$xyz=7(x+y+z)$



${xyzover 7}=x+y+z$



Where $x,y,z$ are prime numbers, so one of them must be $7$.



For example $x$



$yz=(7+y+z)$



Now I have to find these two numbers



I have found $5$ and $3$



$3*5*7=7(15)$



Which is correct.



My question is; Are there any other prime numbers, which
meet the equation?






prime-numbers systems-of-equations




prime-numbers systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 12:34









Henrik

5,99792030




5,99792030










asked Dec 2 '18 at 12:17









KukozKukoz

278




278












  • $begingroup$
    We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:21












  • $begingroup$
    Yeah, I wrote it, but a little differently ;)
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:23










  • $begingroup$
    I don't see where you write that we are done, not even a little differently:)
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:24












  • $begingroup$
    If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:26








  • 2




    $begingroup$
    $yz=y+z+7$ is the same as $(y-1)(z-1)=8$
    $endgroup$
    – Empy2
    Dec 2 '18 at 12:36


















  • $begingroup$
    We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:21












  • $begingroup$
    Yeah, I wrote it, but a little differently ;)
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:23










  • $begingroup$
    I don't see where you write that we are done, not even a little differently:)
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 12:24












  • $begingroup$
    If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
    $endgroup$
    – Kukoz
    Dec 2 '18 at 12:26








  • 2




    $begingroup$
    $yz=y+z+7$ is the same as $(y-1)(z-1)=8$
    $endgroup$
    – Empy2
    Dec 2 '18 at 12:36
















$begingroup$
We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 12:21






$begingroup$
We have $7mid xyz$, so that one of the primes $x,y,z$ is $7$. Without loss of generality, $x=7$. Then $yz=7+y+z$. However, very quickly $yz$ is bigger than $7+y+z$, and we are done.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 12:21














$begingroup$
Yeah, I wrote it, but a little differently ;)
$endgroup$
– Kukoz
Dec 2 '18 at 12:23




$begingroup$
Yeah, I wrote it, but a little differently ;)
$endgroup$
– Kukoz
Dec 2 '18 at 12:23












$begingroup$
I don't see where you write that we are done, not even a little differently:)
$endgroup$
– Dietrich Burde
Dec 2 '18 at 12:24






$begingroup$
I don't see where you write that we are done, not even a little differently:)
$endgroup$
– Dietrich Burde
Dec 2 '18 at 12:24














$begingroup$
If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
$endgroup$
– Kukoz
Dec 2 '18 at 12:26






$begingroup$
If " ${xyzover 7}=x+y+z $ Where x,y,z is prime number, so one of them must be 7." is not the same as " We have $ 7mid xyz $ so that one of the primes x,y,z is 7." I would be blind...
$endgroup$
– Kukoz
Dec 2 '18 at 12:26






2




2




$begingroup$
$yz=y+z+7$ is the same as $(y-1)(z-1)=8$
$endgroup$
– Empy2
Dec 2 '18 at 12:36




$begingroup$
$yz=y+z+7$ is the same as $(y-1)(z-1)=8$
$endgroup$
– Empy2
Dec 2 '18 at 12:36










1 Answer
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$begingroup$

You have concluded that we may assume that $pq=7+p+q$ and $ple q$ for primes $p,q$. For $pge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.






share|cite|improve this answer









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    $begingroup$

    You have concluded that we may assume that $pq=7+p+q$ and $ple q$ for primes $p,q$. For $pge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have concluded that we may assume that $pq=7+p+q$ and $ple q$ for primes $p,q$. For $pge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have concluded that we may assume that $pq=7+p+q$ and $ple q$ for primes $p,q$. For $pge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.






        share|cite|improve this answer









        $endgroup$



        You have concluded that we may assume that $pq=7+p+q$ and $ple q$ for primes $p,q$. For $pge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 12:32









        Dietrich BurdeDietrich Burde

        78.3k64386




        78.3k64386






























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