Proof Verification: Prove that $gcd((a+b)^m, (a-b)^m)leq2^m$ for relatively prime $a$ and $b$.












2












$begingroup$


I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:




If $a$ and $b$ are relatively prime integers, then prove that $$gcd((a+b)^m, (a-b)^m)leq2^m$$




This is my attempt:



Let $gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)Rightarrow k|2a$ and $k|2b$.



Now either $k|2$, $k|a$, or $k|b$. If the first is true then $kleq2$, and if the latter two are true then $k|gcd(a,b) Rightarrow k|1$, in which case we can again say that $k=1leq2$.



Now we know that $gcd((a+b)^m,(a-b)^m)=k^m$, and since $kleq2$ it implies $k^mleq2^m$, which proves our proposition. QED.




Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $kleq2$ the most.






elementary-number-theory proof-verification







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$endgroup$












  • $begingroup$
    Just out of curiosity, how do you add the yellow box around questions?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:40
















2












$begingroup$


I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:




If $a$ and $b$ are relatively prime integers, then prove that $$gcd((a+b)^m, (a-b)^m)leq2^m$$




This is my attempt:



Let $gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)Rightarrow k|2a$ and $k|2b$.



Now either $k|2$, $k|a$, or $k|b$. If the first is true then $kleq2$, and if the latter two are true then $k|gcd(a,b) Rightarrow k|1$, in which case we can again say that $k=1leq2$.



Now we know that $gcd((a+b)^m,(a-b)^m)=k^m$, and since $kleq2$ it implies $k^mleq2^m$, which proves our proposition. QED.




Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $kleq2$ the most.






elementary-number-theory proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just out of curiosity, how do you add the yellow box around questions?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:40














2












2








2





$begingroup$


I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:




If $a$ and $b$ are relatively prime integers, then prove that $$gcd((a+b)^m, (a-b)^m)leq2^m$$




This is my attempt:



Let $gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)Rightarrow k|2a$ and $k|2b$.



Now either $k|2$, $k|a$, or $k|b$. If the first is true then $kleq2$, and if the latter two are true then $k|gcd(a,b) Rightarrow k|1$, in which case we can again say that $k=1leq2$.



Now we know that $gcd((a+b)^m,(a-b)^m)=k^m$, and since $kleq2$ it implies $k^mleq2^m$, which proves our proposition. QED.




Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $kleq2$ the most.






elementary-number-theory proof-verification







share|cite|improve this question











$endgroup$




I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:




If $a$ and $b$ are relatively prime integers, then prove that $$gcd((a+b)^m, (a-b)^m)leq2^m$$




This is my attempt:



Let $gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)Rightarrow k|2a$ and $k|2b$.



Now either $k|2$, $k|a$, or $k|b$. If the first is true then $kleq2$, and if the latter two are true then $k|gcd(a,b) Rightarrow k|1$, in which case we can again say that $k=1leq2$.



Now we know that $gcd((a+b)^m,(a-b)^m)=k^m$, and since $kleq2$ it implies $k^mleq2^m$, which proves our proposition. QED.




Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $kleq2$ the most.






elementary-number-theory proof-verification




elementary-number-theory proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 12:37









greedoid

42.2k1152105




42.2k1152105










asked Jan 18 at 12:04









Naman KumarNaman Kumar

19613




19613












  • $begingroup$
    Just out of curiosity, how do you add the yellow box around questions?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:40


















  • $begingroup$
    Just out of curiosity, how do you add the yellow box around questions?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:40
















$begingroup$
Just out of curiosity, how do you add the yellow box around questions?
$endgroup$
– Naman Kumar
Jan 18 at 12:40




$begingroup$
Just out of curiosity, how do you add the yellow box around questions?
$endgroup$
– Naman Kumar
Jan 18 at 12:40










1 Answer
1






active

oldest

votes


















3












$begingroup$

This is not correct: Now either $k|2$, $k|a$, or $k|b$.



It is better if you take prime $pmid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.



So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $lgeq 2$ then $4mid a+b$ and $4mid a-b$ so $4mid 2a$ so $2mid a$ and the same is true for $b$, that is $2mid b$. Thus $lleq 1$ and so $k=2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:24










  • $begingroup$
    Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
    $endgroup$
    – greedoid
    Jan 18 at 12:25








  • 2




    $begingroup$
    Oh, that makes a lot more sense now. Thank you.
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:31











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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3












$begingroup$

This is not correct: Now either $k|2$, $k|a$, or $k|b$.



It is better if you take prime $pmid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.



So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $lgeq 2$ then $4mid a+b$ and $4mid a-b$ so $4mid 2a$ so $2mid a$ and the same is true for $b$, that is $2mid b$. Thus $lleq 1$ and so $k=2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:24










  • $begingroup$
    Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
    $endgroup$
    – greedoid
    Jan 18 at 12:25








  • 2




    $begingroup$
    Oh, that makes a lot more sense now. Thank you.
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:31
















3












$begingroup$

This is not correct: Now either $k|2$, $k|a$, or $k|b$.



It is better if you take prime $pmid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.



So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $lgeq 2$ then $4mid a+b$ and $4mid a-b$ so $4mid 2a$ so $2mid a$ and the same is true for $b$, that is $2mid b$. Thus $lleq 1$ and so $k=2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:24










  • $begingroup$
    Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
    $endgroup$
    – greedoid
    Jan 18 at 12:25








  • 2




    $begingroup$
    Oh, that makes a lot more sense now. Thank you.
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:31














3












3








3





$begingroup$

This is not correct: Now either $k|2$, $k|a$, or $k|b$.



It is better if you take prime $pmid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.



So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $lgeq 2$ then $4mid a+b$ and $4mid a-b$ so $4mid 2a$ so $2mid a$ and the same is true for $b$, that is $2mid b$. Thus $lleq 1$ and so $k=2$.






share|cite|improve this answer











$endgroup$



This is not correct: Now either $k|2$, $k|a$, or $k|b$.



It is better if you take prime $pmid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.



So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $lgeq 2$ then $4mid a+b$ and $4mid a-b$ so $4mid 2a$ so $2mid a$ and the same is true for $b$, that is $2mid b$. Thus $lleq 1$ and so $k=2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 12:32

























answered Jan 18 at 12:15









greedoidgreedoid

42.2k1152105




42.2k1152105












  • $begingroup$
    Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:24










  • $begingroup$
    Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
    $endgroup$
    – greedoid
    Jan 18 at 12:25








  • 2




    $begingroup$
    Oh, that makes a lot more sense now. Thank you.
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:31


















  • $begingroup$
    Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:24










  • $begingroup$
    Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
    $endgroup$
    – greedoid
    Jan 18 at 12:25








  • 2




    $begingroup$
    Oh, that makes a lot more sense now. Thank you.
    $endgroup$
    – Naman Kumar
    Jan 18 at 12:31
















$begingroup$
Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
$endgroup$
– Naman Kumar
Jan 18 at 12:24




$begingroup$
Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p mid k$ instead of simply solving using $k$. Can you please elaborate?
$endgroup$
– Naman Kumar
Jan 18 at 12:24












$begingroup$
Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
$endgroup$
– greedoid
Jan 18 at 12:25






$begingroup$
Say $10 mid 2cdot 5cdot 2$ where $a=5$, $b=2$
$endgroup$
– greedoid
Jan 18 at 12:25






2




2




$begingroup$
Oh, that makes a lot more sense now. Thank you.
$endgroup$
– Naman Kumar
Jan 18 at 12:31




$begingroup$
Oh, that makes a lot more sense now. Thank you.
$endgroup$
– Naman Kumar
Jan 18 at 12:31


















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