Congruence equation for distinct primes $p,q$.
Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$
This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.
Proof:
Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$
Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$
Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
$square$
elementary-number-theory proof-verification prime-numbers alternative-proof
edited Jan 4 at 21:50
greedoid
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
add a comment |
Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$
This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.
Proof:
Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$
Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$
Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
$square$
elementary-number-theory proof-verification prime-numbers alternative-proof
edited Jan 4 at 21:50
greedoid
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
add a comment |
Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$
This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.
Proof:
Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$
Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$
Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
$square$
elementary-number-theory proof-verification prime-numbers alternative-proof
edited Jan 4 at 21:50
greedoid
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$
This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.
Proof:
Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$
Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$
Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
$square$
elementary-number-theory proof-verification prime-numbers alternative-proof
elementary-number-theory proof-verification prime-numbers alternative-proof
edited Jan 4 at 21:50
greedoid
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
edited Jan 4 at 21:50
greedoid
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
edited Jan 4 at 21:50
greedoid
38.6k114797
edited Jan 4 at 21:50
greedoid
38.6k114797
edited Jan 4 at 21:50
greedoid
38.6k114797
38.6k114797
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
asked Sep 3 '18 at 9:02
Moed Pol BolloMoed Pol Bollo
4231310
4231310
add a comment |
add a comment |
1 Answer
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Your prof is correct.
Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
and thus a conclusion.
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
add a comment |
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Your prof is correct.
Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
and thus a conclusion.
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
add a comment |
Your prof is correct.
Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
and thus a conclusion.
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
add a comment |
Your prof is correct.
Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
and thus a conclusion.
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
Your prof is correct.
Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
and thus a conclusion.
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
answered Sep 3 '18 at 9:08
greedoidgreedoid
38.6k114797
38.6k114797
add a comment |
add a comment |
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