Prove that number $4p^2+1$ can show as sum of squares of three different numbers [duplicate]
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Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers
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Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k in mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?
discrete-mathematics modular-arithmetic divisibility
asked Nov 21 at 16:23
Marko Škorić
69810
marked as duplicate by Steven Stadnicki, vrugtehagel, amWhy discrete-mathematics
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Nov 21 at 19:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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This question already has an answer here:
Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers
2 answers
Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k in mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?
discrete-mathematics modular-arithmetic divisibility
asked Nov 21 at 16:23
Marko Škorić
69810
marked as duplicate by Steven Stadnicki, vrugtehagel, amWhy discrete-mathematics
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Nov 21 at 19:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
1
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
no you can not include 0
– Marko Škorić
Nov 21 at 16:48
add a comment |
up vote
3
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up vote
3
down vote
favorite
This question already has an answer here:
Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers
2 answers
Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k in mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?
discrete-mathematics modular-arithmetic divisibility
asked Nov 21 at 16:23
Marko Škorić
69810
This question already has an answer here:
Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers
2 answers
Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k in mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?
This question already has an answer here:
Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers
2 answers
discrete-mathematics modular-arithmetic divisibility
discrete-mathematics modular-arithmetic divisibility
asked Nov 21 at 16:23
Marko Škorić
69810
asked Nov 21 at 16:23
Marko Škorić
69810
asked Nov 21 at 16:23
Marko Škorić
69810
asked Nov 21 at 16:23
Marko Škorić
69810
asked Nov 21 at 16:23
Marko Škorić
69810
69810
marked as duplicate by Steven Stadnicki, vrugtehagel, amWhy discrete-mathematics
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Nov 21 at 19:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Steven Stadnicki, vrugtehagel, amWhy discrete-mathematics
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Nov 21 at 19:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
1
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
no you can not include 0
– Marko Škorić
Nov 21 at 16:48
add a comment |
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
1
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
no you can not include 0
– Marko Škorić
Nov 21 at 16:48
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
1
1
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
no you can not include 0
– Marko Škorić
Nov 21 at 16:48
no you can not include 0
– Marko Škorić
Nov 21 at 16:48
add a comment |
1 Answer
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2
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accepted
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
answered Nov 21 at 16:46
Raptor
1,3612425
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
answered Nov 21 at 16:46
Raptor
1,3612425
add a comment |
up vote
2
down vote
accepted
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
answered Nov 21 at 16:46
Raptor
1,3612425
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
answered Nov 21 at 16:46
Raptor
1,3612425
Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$therefore 144=a^2+b^2+c^2 qquad$ and $qquad -48=-(2b+4c)$
So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$
answered Nov 21 at 16:46
Raptor
1,3612425
edited Nov 21 at 16:51
answered Nov 21 at 16:46
Raptor
1,3612425
answered Nov 21 at 16:46
Raptor
1,3612425
answered Nov 21 at 16:46
Raptor
1,3612425
1,3612425
add a comment |
add a comment |
Does $0$ count as a square here? I mean $4cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $pneq 0$.
– Snookie
Nov 21 at 16:42
1
It is possible to ignore $0$, as I did in my answer @Snookie .
– Raptor
Nov 21 at 16:46
no you can not include 0
– Marko Škorić
Nov 21 at 16:48