if $n = a_1a_2 cdots a_r + 2$, then $a_i nmid n$ for each integer $i (1 leq i leq r)$.
$begingroup$
Let $a_1, a_2, cdots , a_r$ be odd integers where $a_i > 1$ for $i = 1, 2, cdots , r$. Prove that if $n = a_1a_2 cdots a_r + 2$, then $a_i nmid n$ for each integer $i (1 leq i leq r)$.
Let $a_i mid n$, then $a_i mid a_1a_2 cdots a_r + 2$ also $a_i mid a_1a_2 cdots a_r$, thus $a_i mid a_1a_2 cdots a_r+2 - a_1a_2 cdots a_r implies a_i mid 2$, which is not possible since $a_i > 1$ is odd integer.
Is the logic correct?
number-theory proof-verification divisibility integers
edited Nov 28 '18 at 6:59
user26857
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
$endgroup$
add a comment |
$begingroup$
Let $a_1, a_2, cdots , a_r$ be odd integers where $a_i > 1$ for $i = 1, 2, cdots , r$. Prove that if $n = a_1a_2 cdots a_r + 2$, then $a_i nmid n$ for each integer $i (1 leq i leq r)$.
Let $a_i mid n$, then $a_i mid a_1a_2 cdots a_r + 2$ also $a_i mid a_1a_2 cdots a_r$, thus $a_i mid a_1a_2 cdots a_r+2 - a_1a_2 cdots a_r implies a_i mid 2$, which is not possible since $a_i > 1$ is odd integer.
Is the logic correct?
number-theory proof-verification divisibility integers
edited Nov 28 '18 at 6:59
user26857
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
$endgroup$
$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38
add a comment |
$begingroup$
Let $a_1, a_2, cdots , a_r$ be odd integers where $a_i > 1$ for $i = 1, 2, cdots , r$. Prove that if $n = a_1a_2 cdots a_r + 2$, then $a_i nmid n$ for each integer $i (1 leq i leq r)$.
Let $a_i mid n$, then $a_i mid a_1a_2 cdots a_r + 2$ also $a_i mid a_1a_2 cdots a_r$, thus $a_i mid a_1a_2 cdots a_r+2 - a_1a_2 cdots a_r implies a_i mid 2$, which is not possible since $a_i > 1$ is odd integer.
Is the logic correct?
number-theory proof-verification divisibility integers
edited Nov 28 '18 at 6:59
user26857
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
$endgroup$
Let $a_1, a_2, cdots , a_r$ be odd integers where $a_i > 1$ for $i = 1, 2, cdots , r$. Prove that if $n = a_1a_2 cdots a_r + 2$, then $a_i nmid n$ for each integer $i (1 leq i leq r)$.
Let $a_i mid n$, then $a_i mid a_1a_2 cdots a_r + 2$ also $a_i mid a_1a_2 cdots a_r$, thus $a_i mid a_1a_2 cdots a_r+2 - a_1a_2 cdots a_r implies a_i mid 2$, which is not possible since $a_i > 1$ is odd integer.
Is the logic correct?
number-theory proof-verification divisibility integers
number-theory proof-verification divisibility integers
edited Nov 28 '18 at 6:59
user26857
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
edited Nov 28 '18 at 6:59
user26857
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
edited Nov 28 '18 at 6:59
user26857
39.4k124183
edited Nov 28 '18 at 6:59
user26857
39.4k124183
edited Nov 28 '18 at 6:59
user26857
39.4k124183
39.4k124183
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
asked Nov 28 '18 at 6:11
user8795user8795
5,66462047
5,66462047
$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38
add a comment |
$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38
$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38
$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes, your reasoning is sound and correct.
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
$endgroup$
add a comment |
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$begingroup$
Yes, your reasoning is sound and correct.
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
$endgroup$
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$begingroup$
Yes, your reasoning is sound and correct.
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Yes, your reasoning is sound and correct.
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
$endgroup$
Yes, your reasoning is sound and correct.
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
answered Nov 28 '18 at 6:12
plattyplatty
3,370320
3,370320
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$begingroup$
Yes. Alternatively by Euclid $,gcd(n,a_i) = gcd(2,a_i) $ by $ 2 = nbmod a_i $
$endgroup$
– Bill Dubuque
Nov 28 '18 at 19:38