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Prove that it does not exist two number such that $m^2+n^2=6 underbrace {0 cdots 0}$
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Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?
modular-arithmetic divisibility
asked Nov 13 at 4:51
Marko Škorić
58110
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Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?
modular-arithmetic divisibility
asked Nov 13 at 4:51
Marko Škorić
58110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?
modular-arithmetic divisibility
asked Nov 13 at 4:51
Marko Škorić
58110
Prove that it does not exist a two number $m,nin mathbb N$ such that $m^2+n^2=6 underbrace {0 cdots 0}$, in solution he choose to divide by 9, but i do not know why that number, I know that remainder of $m^2 equiv r_1(mod 9)$ and $n^2 equiv r_2(mod 9)$ $r_1,r_2 in {0,7,1,4}$ $m^2+n^2equiv r_3(mod 9)$ $r3in {1,2,4,5,7,8}$ but $10equiv1 (mod 9)$ $10^{13} equiv 1 (mod9)$ $6cdot 10^{13} equiv 6 (mod9)$ so this is not true, but they choose a number nine because it is easy to prove or there is something else?
modular-arithmetic divisibility
modular-arithmetic divisibility
asked Nov 13 at 4:51
Marko Škorić
58110
asked Nov 13 at 4:51
Marko Škorić
58110
edited Nov 13 at 5:00
asked Nov 13 at 4:51
Marko Škorić
58110
asked Nov 13 at 4:51
Marko Škorić
58110
asked Nov 13 at 4:51
Marko Škorić
58110
58110
add a comment |
add a comment |
2 Answers
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$60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.
answered Nov 13 at 4:57
Robert Israel
313k23206452
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Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
Prime factorizaton of $6underbrace {0cdots 0}$ is
$$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$
Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers
answered 17 hours ago
Samurai
937
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
$60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.
answered Nov 13 at 4:57
Robert Israel
313k23206452
add a comment |
up vote
5
down vote
$60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.
answered Nov 13 at 4:57
Robert Israel
313k23206452
add a comment |
up vote
5
down vote
up vote
5
down vote
$60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.
answered Nov 13 at 4:57
Robert Israel
313k23206452
$60ldots 0 = 6 times 10^n$ for some $n$. The point is that $10 equiv 1 mod 9$, so $6 times 10^n equiv 6 mod 9$. And thus it suffices to show that the sum of two squares can't be congruent to $6 mod 9$.
answered Nov 13 at 4:57
Robert Israel
313k23206452
answered Nov 13 at 4:57
Robert Israel
313k23206452
answered Nov 13 at 4:57
Robert Israel
313k23206452
answered Nov 13 at 4:57
Robert Israel
313k23206452
313k23206452
add a comment |
add a comment |
up vote
0
down vote
Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
Prime factorizaton of $6underbrace {0cdots 0}$ is
$$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$
Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers
answered 17 hours ago
Samurai
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
Prime factorizaton of $6underbrace {0cdots 0}$ is
$$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$
Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers
answered 17 hours ago
Samurai
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
Prime factorizaton of $6underbrace {0cdots 0}$ is
$$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$
Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers
answered 17 hours ago
Samurai
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Any number ,say $n$, is expressible as sum of two squares if it either has no prime divisor of the form ${4k}+3$ or if it has prime divisor of the form $4k+3$ then its power is even.
Prime factorizaton of $6underbrace {0cdots 0}$ is
$$6underbrace {0cdots 0}= {3}times {2^n}times {5^m}$$
Since the power of $3$ ,which is a prime of the form $4k+3$ ,is $1$ which is an odd number, $6underbrace {0cdots 0}$ cannot be expressed as sum of two integers
answered 17 hours ago
Samurai
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
answered 17 hours ago
Samurai
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 17 hours ago
Samurai
937
answered 17 hours ago
Samurai
937
937
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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