If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$ [closed]
0
$begingroup$
Can you help me prove this number theory claim:
If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$
Thanks
number-theory
edited Dec 22 '18 at 15:15
amWhy
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
$endgroup$
closed as off-topic by amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost Dec 23 '18 at 16:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
Can you help me prove this number theory claim:
If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$
Thanks
number-theory
edited Dec 22 '18 at 15:15
amWhy
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
$endgroup$
closed as off-topic by amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost Dec 23 '18 at 16:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
2
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50
add a comment |
0
0
0
$begingroup$
Can you help me prove this number theory claim:
If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$
Thanks
number-theory
edited Dec 22 '18 at 15:15
amWhy
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
$endgroup$
Can you help me prove this number theory claim:
If p is a prime, p divides $a^2+b^2$ and p divides $a^2$, then p divides $b^2$
Thanks
number-theory
number-theory
edited Dec 22 '18 at 15:15
amWhy
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
edited Dec 22 '18 at 15:15
amWhy
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
edited Dec 22 '18 at 15:15
amWhy
1
edited Dec 22 '18 at 15:15
amWhy
1
edited Dec 22 '18 at 15:15
amWhy
1
1
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
asked May 19 '17 at 17:45
Pablo MelloPablo Mello
526
526
closed as off-topic by amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost Dec 23 '18 at 16:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost Dec 23 '18 at 16:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, mrtaurho, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
2
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50
add a comment |
3
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
2
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50
3
3
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
2
2
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
0
$begingroup$
$p|a^2+b^2-a^2=b^2$
Especially $p|b$
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
2
$begingroup$
Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
2
2
2
$begingroup$
Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
$endgroup$
Badjohn is right: If $kp=a^2+b^2$ and $mp=a^2$ for some integers $k$ and $m$, then $(k-m)p=b^2$.
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
answered May 19 '17 at 17:49
WuestenfuxWuestenfux
4,2871413
4,2871413
add a comment |
add a comment |
0
$begingroup$
$p|a^2+b^2-a^2=b^2$
Especially $p|b$
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
0
$begingroup$
$p|a^2+b^2-a^2=b^2$
Especially $p|b$
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
0
0
0
$begingroup$
$p|a^2+b^2-a^2=b^2$
Especially $p|b$
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
$p|a^2+b^2-a^2=b^2$
Especially $p|b$
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
answered May 19 '17 at 17:49
Marios GretsasMarios Gretsas
8,48011437
8,48011437
add a comment |
add a comment |
3
$begingroup$
The squares are maybe there just to add some confusion. Try ignoring them for a moment and see whether the question looks easier.
$endgroup$
– badjohn
May 19 '17 at 17:46
2
$begingroup$
You don't even need $p$ to be prime. It's all red herrings.
$endgroup$
– Arthur
May 19 '17 at 17:48
$begingroup$
@badjohn That's a good point, thanks!
$endgroup$
– Pablo Mello
May 19 '17 at 17:50