Proof Verification: Prove that $gcd((a+b)^m, (a-b)^m)leq2^m$ for relatively prime $a$ and $b$.
$begingroup$
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
edited Jan 18 at 12:37
greedoid
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
$endgroup$
add a comment |
$begingroup$
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
edited Jan 18 at 12:37
greedoid
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
$endgroup$
$begingroup$
Just out of curiosity, how do you add the yellow box around questions?
$endgroup$
– Naman Kumar
Jan 18 at 12:40
add a comment |
$begingroup$
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
edited Jan 18 at 12:37
greedoid
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
$endgroup$
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
edited Jan 18 at 12:37
greedoid
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
edited Jan 18 at 12:37
greedoid
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
edited Jan 18 at 12:37
greedoid
42.2k1152105
edited Jan 18 at 12:37
greedoid
42.2k1152105
edited Jan 18 at 12:37
greedoid
42.2k1152105
42.2k1152105
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
asked Jan 18 at 12:04
Naman KumarNaman Kumar
19613
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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votes
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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 $k=2$.
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 $k=2$.
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $k=2$.
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $k=2$.
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
$endgroup$
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 $k=2$.
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
edited Jan 18 at 12:32
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
answered Jan 18 at 12:15
greedoidgreedoid
42.2k1152105
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Just out of curiosity, how do you add the yellow box around questions?
$endgroup$
– Naman Kumar
Jan 18 at 12:40