How to solve this number theory question from INMO?












6












$begingroup$



Solve the following equation in integers:



$$m(4m^2 + m + 12) = 3(p^n -1)$$



where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.




This simplifies to :



$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$



$gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.



Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7text{ or }49$$



This means the prime $p$ is 7. and $4m + 1 = 3*7^ktext{ or }7^k$



If $gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 leq 3*7^2 < 7^3 leq 4m + 1$$



How do we arrive at the above inequality chain?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Using the exact same reasoning as in the previous paragraph.
    $endgroup$
    – Peter Taylor
    Dec 25 '18 at 7:21
















6












$begingroup$



Solve the following equation in integers:



$$m(4m^2 + m + 12) = 3(p^n -1)$$



where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.




This simplifies to :



$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$



$gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.



Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7text{ or }49$$



This means the prime $p$ is 7. and $4m + 1 = 3*7^ktext{ or }7^k$



If $gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 leq 3*7^2 < 7^3 leq 4m + 1$$



How do we arrive at the above inequality chain?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Using the exact same reasoning as in the previous paragraph.
    $endgroup$
    – Peter Taylor
    Dec 25 '18 at 7:21














6












6








6





$begingroup$



Solve the following equation in integers:



$$m(4m^2 + m + 12) = 3(p^n -1)$$



where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.




This simplifies to :



$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$



$gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.



Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7text{ or }49$$



This means the prime $p$ is 7. and $4m + 1 = 3*7^ktext{ or }7^k$



If $gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 leq 3*7^2 < 7^3 leq 4m + 1$$



How do we arrive at the above inequality chain?






elementary-number-theory







share|cite|improve this question











$endgroup$





Solve the following equation in integers:



$$m(4m^2 + m + 12) = 3(p^n -1)$$



where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.




This simplifies to :



$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$



$gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.



Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7text{ or }49$$



This means the prime $p$ is 7. and $4m + 1 = 3*7^ktext{ or }7^k$



If $gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 leq 3*7^2 < 7^3 leq 4m + 1$$



How do we arrive at the above inequality chain?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 16:54









greedoid

42k1152105




42k1152105










asked Dec 25 '18 at 7:05









saisanjeevsaisanjeev

987212




987212








  • 2




    $begingroup$
    Using the exact same reasoning as in the previous paragraph.
    $endgroup$
    – Peter Taylor
    Dec 25 '18 at 7:21














  • 2




    $begingroup$
    Using the exact same reasoning as in the previous paragraph.
    $endgroup$
    – Peter Taylor
    Dec 25 '18 at 7:21








2




2




$begingroup$
Using the exact same reasoning as in the previous paragraph.
$endgroup$
– Peter Taylor
Dec 25 '18 at 7:21




$begingroup$
Using the exact same reasoning as in the previous paragraph.
$endgroup$
– Peter Taylor
Dec 25 '18 at 7:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since $m^2+3 =7^l$ or $m^2+3= 3cdot 7^l$ we see that $m^2+3 leq 3cdot 7^l$



Now we have $7^2mid m^2+3$ and $7^3not{mid} ;m^2+3$ so $l=2$.



And since $7^3mid 4m+1$ we have $7^3leq 4m+1$. Clearly $3cdot 7^2<7^3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, you don't like the answer?
    $endgroup$
    – greedoid
    Jan 4 at 21:44










  • $begingroup$
    Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
    $endgroup$
    – caffeinemachine
    Jan 26 at 16:26










  • $begingroup$
    There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
    $endgroup$
    – greedoid
    Jan 26 at 21:52











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1 Answer
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1












$begingroup$

Since $m^2+3 =7^l$ or $m^2+3= 3cdot 7^l$ we see that $m^2+3 leq 3cdot 7^l$



Now we have $7^2mid m^2+3$ and $7^3not{mid} ;m^2+3$ so $l=2$.



And since $7^3mid 4m+1$ we have $7^3leq 4m+1$. Clearly $3cdot 7^2<7^3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, you don't like the answer?
    $endgroup$
    – greedoid
    Jan 4 at 21:44










  • $begingroup$
    Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
    $endgroup$
    – caffeinemachine
    Jan 26 at 16:26










  • $begingroup$
    There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
    $endgroup$
    – greedoid
    Jan 26 at 21:52
















1












$begingroup$

Since $m^2+3 =7^l$ or $m^2+3= 3cdot 7^l$ we see that $m^2+3 leq 3cdot 7^l$



Now we have $7^2mid m^2+3$ and $7^3not{mid} ;m^2+3$ so $l=2$.



And since $7^3mid 4m+1$ we have $7^3leq 4m+1$. Clearly $3cdot 7^2<7^3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, you don't like the answer?
    $endgroup$
    – greedoid
    Jan 4 at 21:44










  • $begingroup$
    Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
    $endgroup$
    – caffeinemachine
    Jan 26 at 16:26










  • $begingroup$
    There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
    $endgroup$
    – greedoid
    Jan 26 at 21:52














1












1








1





$begingroup$

Since $m^2+3 =7^l$ or $m^2+3= 3cdot 7^l$ we see that $m^2+3 leq 3cdot 7^l$



Now we have $7^2mid m^2+3$ and $7^3not{mid} ;m^2+3$ so $l=2$.



And since $7^3mid 4m+1$ we have $7^3leq 4m+1$. Clearly $3cdot 7^2<7^3$.






share|cite|improve this answer









$endgroup$



Since $m^2+3 =7^l$ or $m^2+3= 3cdot 7^l$ we see that $m^2+3 leq 3cdot 7^l$



Now we have $7^2mid m^2+3$ and $7^3not{mid} ;m^2+3$ so $l=2$.



And since $7^3mid 4m+1$ we have $7^3leq 4m+1$. Clearly $3cdot 7^2<7^3$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 16:46









greedoidgreedoid

42k1152105




42k1152105












  • $begingroup$
    So, you don't like the answer?
    $endgroup$
    – greedoid
    Jan 4 at 21:44










  • $begingroup$
    Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
    $endgroup$
    – caffeinemachine
    Jan 26 at 16:26










  • $begingroup$
    There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
    $endgroup$
    – greedoid
    Jan 26 at 21:52


















  • $begingroup$
    So, you don't like the answer?
    $endgroup$
    – greedoid
    Jan 4 at 21:44










  • $begingroup$
    Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
    $endgroup$
    – caffeinemachine
    Jan 26 at 16:26










  • $begingroup$
    There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
    $endgroup$
    – greedoid
    Jan 26 at 21:52
















$begingroup$
So, you don't like the answer?
$endgroup$
– greedoid
Jan 4 at 21:44




$begingroup$
So, you don't like the answer?
$endgroup$
– greedoid
Jan 4 at 21:44












$begingroup$
Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
$endgroup$
– caffeinemachine
Jan 26 at 16:26




$begingroup$
Hey greedoid. I think you have a great answer. I have seen that you are generally very adept at solving contest math questions (and I am sure higher math questions also). Can you give me some tips for improvement. Thanks.
$endgroup$
– caffeinemachine
Jan 26 at 16:26












$begingroup$
There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
$endgroup$
– greedoid
Jan 26 at 21:52




$begingroup$
There is nothing new I can say to you. Try to solve as much problems of different types and read other solutions. Also read articles on stuff usualy apperas on contest....
$endgroup$
– greedoid
Jan 26 at 21:52


















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