Prove that if $m, n, dinmathbb{Z^+}$ and $d | m^2n+1, d | mn^2+1$ then $d | m^3+1, d | n^3+1$
$begingroup$
Prove that if $m, n, dinmathbb{Z^+}$ and $d | m^2n+1, d | mn^2+1$ then $d | m^3+1, d | n^3+1$
My attempt:
So I said that from this we can safely say that
$$m^2nequiv-1pmod{d}, mn^2equiv-1pmod{d}$$
This means that:
$$mn(m)equiv mn(n)pmod{d}$$
Now if $gcd(mn, d)=1$ then we will get that
$$mequiv{n}pmod{d}$$
Now in our original equation we will get
$$m^2n+1equiv0pmod{d}, mn^2+1equiv0pmod{d}$$
$$therefore m^3+1equiv0pmod{d}, n^3+1equiv0pmod{d}$$
But what if $gcd(mn,d)gt{1}$, what will I have to do? And is that even possible in the first place? Thank you in advance.
elementary-number-theory
asked Dec 15 '18 at 4:22
user587054user587054
46511
$endgroup$
add a comment |
$begingroup$
Prove that if $m, n, dinmathbb{Z^+}$ and $d | m^2n+1, d | mn^2+1$ then $d | m^3+1, d | n^3+1$
My attempt:
So I said that from this we can safely say that
$$m^2nequiv-1pmod{d}, mn^2equiv-1pmod{d}$$
This means that:
$$mn(m)equiv mn(n)pmod{d}$$
Now if $gcd(mn, d)=1$ then we will get that
$$mequiv{n}pmod{d}$$
Now in our original equation we will get
$$m^2n+1equiv0pmod{d}, mn^2+1equiv0pmod{d}$$
$$therefore m^3+1equiv0pmod{d}, n^3+1equiv0pmod{d}$$
But what if $gcd(mn,d)gt{1}$, what will I have to do? And is that even possible in the first place? Thank you in advance.
elementary-number-theory
asked Dec 15 '18 at 4:22
user587054user587054
46511
$endgroup$
add a comment |
$begingroup$
Prove that if $m, n, dinmathbb{Z^+}$ and $d | m^2n+1, d | mn^2+1$ then $d | m^3+1, d | n^3+1$
My attempt:
So I said that from this we can safely say that
$$m^2nequiv-1pmod{d}, mn^2equiv-1pmod{d}$$
This means that:
$$mn(m)equiv mn(n)pmod{d}$$
Now if $gcd(mn, d)=1$ then we will get that
$$mequiv{n}pmod{d}$$
Now in our original equation we will get
$$m^2n+1equiv0pmod{d}, mn^2+1equiv0pmod{d}$$
$$therefore m^3+1equiv0pmod{d}, n^3+1equiv0pmod{d}$$
But what if $gcd(mn,d)gt{1}$, what will I have to do? And is that even possible in the first place? Thank you in advance.
elementary-number-theory
asked Dec 15 '18 at 4:22
user587054user587054
46511
$endgroup$
Prove that if $m, n, dinmathbb{Z^+}$ and $d | m^2n+1, d | mn^2+1$ then $d | m^3+1, d | n^3+1$
My attempt:
So I said that from this we can safely say that
$$m^2nequiv-1pmod{d}, mn^2equiv-1pmod{d}$$
This means that:
$$mn(m)equiv mn(n)pmod{d}$$
Now if $gcd(mn, d)=1$ then we will get that
$$mequiv{n}pmod{d}$$
Now in our original equation we will get
$$m^2n+1equiv0pmod{d}, mn^2+1equiv0pmod{d}$$
$$therefore m^3+1equiv0pmod{d}, n^3+1equiv0pmod{d}$$
But what if $gcd(mn,d)gt{1}$, what will I have to do? And is that even possible in the first place? Thank you in advance.
elementary-number-theory
elementary-number-theory
asked Dec 15 '18 at 4:22
user587054user587054
46511
asked Dec 15 '18 at 4:22
user587054user587054
46511
asked Dec 15 '18 at 4:22
user587054user587054
46511
asked Dec 15 '18 at 4:22
user587054user587054
46511
asked Dec 15 '18 at 4:22
user587054user587054
46511
46511
add a comment |
add a comment |
1 Answer
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Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k mid d$. Observe that: $p^k mid d mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k nmid (n-m)implies p^k mid mnimplies p mid 1$, contradiction $p$ being a prime. Thus $p^k mid (n-m)$,and this means $p^k mid m^3+1$. Thus $d mid (m^3+1)$, and similarly $d mid (n^3+1)$. Note that if $p^r mid (n-m)$, and $p^s mid mn$ with $r+s = k$, then $p mid m^2n implies p mid 1$, contradiction $p$ being a prime.
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
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add a comment |
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$begingroup$
Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k mid d$. Observe that: $p^k mid d mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k nmid (n-m)implies p^k mid mnimplies p mid 1$, contradiction $p$ being a prime. Thus $p^k mid (n-m)$,and this means $p^k mid m^3+1$. Thus $d mid (m^3+1)$, and similarly $d mid (n^3+1)$. Note that if $p^r mid (n-m)$, and $p^s mid mn$ with $r+s = k$, then $p mid m^2n implies p mid 1$, contradiction $p$ being a prime.
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
$endgroup$
add a comment |
$begingroup$
Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k mid d$. Observe that: $p^k mid d mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k nmid (n-m)implies p^k mid mnimplies p mid 1$, contradiction $p$ being a prime. Thus $p^k mid (n-m)$,and this means $p^k mid m^3+1$. Thus $d mid (m^3+1)$, and similarly $d mid (n^3+1)$. Note that if $p^r mid (n-m)$, and $p^s mid mn$ with $r+s = k$, then $p mid m^2n implies p mid 1$, contradiction $p$ being a prime.
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
$endgroup$
add a comment |
$begingroup$
Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k mid d$. Observe that: $p^k mid d mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k nmid (n-m)implies p^k mid mnimplies p mid 1$, contradiction $p$ being a prime. Thus $p^k mid (n-m)$,and this means $p^k mid m^3+1$. Thus $d mid (m^3+1)$, and similarly $d mid (n^3+1)$. Note that if $p^r mid (n-m)$, and $p^s mid mn$ with $r+s = k$, then $p mid m^2n implies p mid 1$, contradiction $p$ being a prime.
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
$endgroup$
Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k mid d$. Observe that: $p^k mid d mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k nmid (n-m)implies p^k mid mnimplies p mid 1$, contradiction $p$ being a prime. Thus $p^k mid (n-m)$,and this means $p^k mid m^3+1$. Thus $d mid (m^3+1)$, and similarly $d mid (n^3+1)$. Note that if $p^r mid (n-m)$, and $p^s mid mn$ with $r+s = k$, then $p mid m^2n implies p mid 1$, contradiction $p$ being a prime.
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
answered Dec 15 '18 at 5:10
DeepSeaDeepSea
70.9k54487
70.9k54487
add a comment |
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