Mixing
Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.
$begingroup$
Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.
I am not sure where to start for this question, any help would be greatly appreciated, thanks!
divisibility
asked Jan 21 at 4:39
macymacy
305
$endgroup$
add a comment |
$begingroup$
Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.
I am not sure where to start for this question, any help would be greatly appreciated, thanks!
divisibility
asked Jan 21 at 4:39
macymacy
305
$endgroup$
add a comment |
$begingroup$
Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.
I am not sure where to start for this question, any help would be greatly appreciated, thanks!
divisibility
asked Jan 21 at 4:39
macymacy
305
$endgroup$
Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.
I am not sure where to start for this question, any help would be greatly appreciated, thanks!
divisibility
divisibility
asked Jan 21 at 4:39
macymacy
305
asked Jan 21 at 4:39
macymacy
305
asked Jan 21 at 4:39
macymacy
305
asked Jan 21 at 4:39
macymacy
305
asked Jan 21 at 4:39
macymacy
305
305
add a comment |
add a comment |
3 Answers
3
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$begingroup$
Suppose $b-1=ak$, $kinmathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3mid12$!
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
$endgroup$
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$begingroup$
As has been stated in other answers,
$$a mid b - 1 Rightarrow a^2 mid b^2 - 2b + 1$$
as $left(b - 1right)^2 = b^2 - 2b + 1$, so
$$a^2 mid left(b^2 - 2b + 4right) - left(b^2 - 2b + 1right) Rightarrow a^2 mid 3$$
Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a mid 12$.
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
$endgroup$
add a comment |
$begingroup$
Well:
$$a|(b-1)implies a^2|(b^2-2b+1)$$
So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$
Can you continue this?
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
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active
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oldest
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$begingroup$
Suppose $b-1=ak$, $kinmathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3mid12$!
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
$endgroup$
add a comment |
$begingroup$
Suppose $b-1=ak$, $kinmathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3mid12$!
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
$endgroup$
add a comment |
$begingroup$
Suppose $b-1=ak$, $kinmathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3mid12$!
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
$endgroup$
Suppose $b-1=ak$, $kinmathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3mid12$!
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
answered Jan 21 at 4:52
YiFanYiFan
4,4711627
4,4711627
add a comment |
add a comment |
$begingroup$
As has been stated in other answers,
$$a mid b - 1 Rightarrow a^2 mid b^2 - 2b + 1$$
as $left(b - 1right)^2 = b^2 - 2b + 1$, so
$$a^2 mid left(b^2 - 2b + 4right) - left(b^2 - 2b + 1right) Rightarrow a^2 mid 3$$
Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a mid 12$.
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
$endgroup$
add a comment |
$begingroup$
As has been stated in other answers,
$$a mid b - 1 Rightarrow a^2 mid b^2 - 2b + 1$$
as $left(b - 1right)^2 = b^2 - 2b + 1$, so
$$a^2 mid left(b^2 - 2b + 4right) - left(b^2 - 2b + 1right) Rightarrow a^2 mid 3$$
Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a mid 12$.
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
$endgroup$
add a comment |
$begingroup$
As has been stated in other answers,
$$a mid b - 1 Rightarrow a^2 mid b^2 - 2b + 1$$
as $left(b - 1right)^2 = b^2 - 2b + 1$, so
$$a^2 mid left(b^2 - 2b + 4right) - left(b^2 - 2b + 1right) Rightarrow a^2 mid 3$$
Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a mid 12$.
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
$endgroup$
As has been stated in other answers,
$$a mid b - 1 Rightarrow a^2 mid b^2 - 2b + 1$$
as $left(b - 1right)^2 = b^2 - 2b + 1$, so
$$a^2 mid left(b^2 - 2b + 4right) - left(b^2 - 2b + 1right) Rightarrow a^2 mid 3$$
Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a mid 12$.
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
edited Jan 21 at 5:04
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
answered Jan 21 at 4:59
John OmielanJohn Omielan
3,5501215
3,5501215
add a comment |
add a comment |
$begingroup$
Well:
$$a|(b-1)implies a^2|(b^2-2b+1)$$
So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$
Can you continue this?
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
Well:
$$a|(b-1)implies a^2|(b^2-2b+1)$$
So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$
Can you continue this?
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
Well:
$$a|(b-1)implies a^2|(b^2-2b+1)$$
So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$
Can you continue this?
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
$endgroup$
Well:
$$a|(b-1)implies a^2|(b^2-2b+1)$$
So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$
Can you continue this?
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 4:53
Rhys HughesRhys Hughes
6,9571530
6,9571530
add a comment |
add a comment |
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