Mixing
Can a prime of form $4m+1$ be expressed as sum of 2 squares in more than one way?
It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.
How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b in Bbb Z^+, a>b$ in only one way?
This is easy to verify for primes up to quite a large value but I can't find a proof.
elementary-number-theory prime-numbers
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
add a comment |
It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.
How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b in Bbb Z^+, a>b$ in only one way?
This is easy to verify for primes up to quite a large value but I can't find a proof.
elementary-number-theory prime-numbers
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
3
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57
add a comment |
It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.
How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b in Bbb Z^+, a>b$ in only one way?
This is easy to verify for primes up to quite a large value but I can't find a proof.
elementary-number-theory prime-numbers
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.
How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b in Bbb Z^+, a>b$ in only one way?
This is easy to verify for primes up to quite a large value but I can't find a proof.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
asked Nov 20 '18 at 20:21
Mark Fischler
32.4k12250
32.4k12250
Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
3
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57
add a comment |
Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
3
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57
Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
3
3
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57
add a comment |
1 Answer
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Denote by $p$ a prime number of the form $4m+1$. Thus $$pequiv 1pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,dinmathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=bigl(p-{b^2bigr)}{d^2}-{b^2}bigl(p-{d^2bigr)}=pbigl({d^2}-{b^2}bigr)equiv 0pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $pmid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=Bigl({a^2}+{b^2}Bigr)Bigl({c^2}+{d^2}Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $pnmid ({ad+bc})$.
This implies $pmid ({ad-bc})$. Analugously $$a,b,c,d<sqrt{p}$$ which implies that $$-p<ad-bc<p Rightarrow ad=bc$$ Hence $amid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$amid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}bigl({a^2}+{b^2}bigr)Rightarrow x=1$$
$$Rightarrow c=aquad d=b$$
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
add a comment |
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Denote by $p$ a prime number of the form $4m+1$. Thus $$pequiv 1pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,dinmathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=bigl(p-{b^2bigr)}{d^2}-{b^2}bigl(p-{d^2bigr)}=pbigl({d^2}-{b^2}bigr)equiv 0pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $pmid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=Bigl({a^2}+{b^2}Bigr)Bigl({c^2}+{d^2}Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $pnmid ({ad+bc})$.
This implies $pmid ({ad-bc})$. Analugously $$a,b,c,d<sqrt{p}$$ which implies that $$-p<ad-bc<p Rightarrow ad=bc$$ Hence $amid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$amid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}bigl({a^2}+{b^2}bigr)Rightarrow x=1$$
$$Rightarrow c=aquad d=b$$
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
add a comment |
Denote by $p$ a prime number of the form $4m+1$. Thus $$pequiv 1pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,dinmathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=bigl(p-{b^2bigr)}{d^2}-{b^2}bigl(p-{d^2bigr)}=pbigl({d^2}-{b^2}bigr)equiv 0pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $pmid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=Bigl({a^2}+{b^2}Bigr)Bigl({c^2}+{d^2}Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $pnmid ({ad+bc})$.
This implies $pmid ({ad-bc})$. Analugously $$a,b,c,d<sqrt{p}$$ which implies that $$-p<ad-bc<p Rightarrow ad=bc$$ Hence $amid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$amid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}bigl({a^2}+{b^2}bigr)Rightarrow x=1$$
$$Rightarrow c=aquad d=b$$
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
add a comment |
Denote by $p$ a prime number of the form $4m+1$. Thus $$pequiv 1pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,dinmathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=bigl(p-{b^2bigr)}{d^2}-{b^2}bigl(p-{d^2bigr)}=pbigl({d^2}-{b^2}bigr)equiv 0pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $pmid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=Bigl({a^2}+{b^2}Bigr)Bigl({c^2}+{d^2}Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $pnmid ({ad+bc})$.
This implies $pmid ({ad-bc})$. Analugously $$a,b,c,d<sqrt{p}$$ which implies that $$-p<ad-bc<p Rightarrow ad=bc$$ Hence $amid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$amid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}bigl({a^2}+{b^2}bigr)Rightarrow x=1$$
$$Rightarrow c=aquad d=b$$
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
Denote by $p$ a prime number of the form $4m+1$. Thus $$pequiv 1pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,dinmathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=bigl(p-{b^2bigr)}{d^2}-{b^2}bigl(p-{d^2bigr)}=pbigl({d^2}-{b^2}bigr)equiv 0pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $pmid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=Bigl({a^2}+{b^2}Bigr)Bigl({c^2}+{d^2}Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $pnmid ({ad+bc})$.
This implies $pmid ({ad-bc})$. Analugously $$a,b,c,d<sqrt{p}$$ which implies that $$-p<ad-bc<p Rightarrow ad=bc$$ Hence $amid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$amid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}bigl({a^2}+{b^2}bigr)Rightarrow x=1$$
$$Rightarrow c=aquad d=b$$
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
answered Nov 20 '18 at 22:33
Dr. Mathva
919316
919316
add a comment |
add a comment |
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Most elementary texts on number theory prove this.
– Lord Shark the Unknown
Nov 20 '18 at 20:28
@mathnoob $3$ is not of the from $4m+1$
– saulspatz
Nov 20 '18 at 20:38
3
very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
– Will Jagy
Nov 20 '18 at 20:43
@Will Jagy - Thanks for that reference - very enlightening.
– marty cohen
Nov 20 '18 at 20:52
@will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow.
– Mark Fischler
Nov 20 '18 at 20:57