Can you substitute congruences into an expression if the conditions are right?

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When using a congruences can we substitute them into an expression. For example, when we look at Fermat's little theorem we notice that $a^p$ is congruent to $a (textmod p)$ when $a$ is an integer and $p$ is prime, so where ever we see $a^p$ where $a$ is an integer and $p$ is prime, can we simply substitute $a (textmod p)$ for $a^p$?

number-theory

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edited Jul 25 at 18:43

asked Jul 24 at 6:29

Charles Quinn

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up vote
1
down vote

favorite

When using a congruences can we substitute them into an expression. For example, when we look at Fermat's little theorem we notice that $a^p$ is congruent to $a (textmod p)$ when $a$ is an integer and $p$ is prime, so where ever we see $a^p$ where $a$ is an integer and $p$ is prime, can we simply substitute $a (textmod p)$ for $a^p$?

number-theory

share|cite|improve this question

edited Jul 25 at 18:43

asked Jul 24 at 6:29

Charles Quinn

533

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

When using a congruences can we substitute them into an expression. For example, when we look at Fermat's little theorem we notice that $a^p$ is congruent to $a (textmod p)$ when $a$ is an integer and $p$ is prime, so where ever we see $a^p$ where $a$ is an integer and $p$ is prime, can we simply substitute $a (textmod p)$ for $a^p$?

number-theory

share|cite|improve this question

edited Jul 25 at 18:43

asked Jul 24 at 6:29

Charles Quinn

533

When using a congruences can we substitute them into an expression. For example, when we look at Fermat's little theorem we notice that $a^p$ is congruent to $a (textmod p)$ when $a$ is an integer and $p$ is prime, so where ever we see $a^p$ where $a$ is an integer and $p$ is prime, can we simply substitute $a (textmod p)$ for $a^p$?

number-theory

share|cite|improve this question

edited Jul 25 at 18:43

asked Jul 24 at 6:29

Charles Quinn

533

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 18:43

edited Jul 25 at 18:43

edited Jul 25 at 18:43

asked Jul 24 at 6:29

Charles Quinn

533

asked Jul 24 at 6:29

Charles Quinn

533

asked Jul 24 at 6:29

Charles Quinn

533

533

add a comment | 

add a comment | 

3 Answers
3

active

oldest

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up vote
0
down vote



accepted

Yes in general using congruences all algebraic rules hold for for the sum and product and we can substitute them into an expression, notably for

• $Aequiv a mod p$


• $Bequiv b mod p$

we have that

• $A+Bequiv a+b mod p$


• $Acdot Bequiv acdot b mod p$

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answered Jul 24 at 6:36

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
  – Charles Quinn
  Jul 25 at 4:25
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

No. It would be very bad notation in some cases.

For example, if you replace $a^p$ by $apmod p$ in the equation $$a^ptimes5=5a^p$$ then you get $$apmod ptimes5=5apmod p$$ so the RHS would be very ambiguous. Do you mean $5apmod p$ or $5times[apmod p]$?

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I do, then I must be careful.
  – Charles Quinn
  Jul 24 at 22:32
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

So, what about taking the Pythagorean triples as an example. We know for Pythagorean triples $x,y,$ and $z$ are integers in $x^2+y^2=z^2$. By Fermats little theorem, we know

• $a^2equiva (textmod 2)$ where $ainmathbbz$

So, $x^2+y^2=z^2$ yields,

• $x+yequivz (textmod 2)$

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Yes in general using congruences all algebraic rules hold for for the sum and product and we can substitute them into an expression, notably for

• $Aequiv a mod p$


• $Bequiv b mod p$

we have that

• $A+Bequiv a+b mod p$


• $Acdot Bequiv acdot b mod p$

share|cite|improve this answer

answered Jul 24 at 6:36

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
  – Charles Quinn
  Jul 25 at 4:25
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Yes in general using congruences all algebraic rules hold for for the sum and product and we can substitute them into an expression, notably for

• $Aequiv a mod p$


• $Bequiv b mod p$

we have that

• $A+Bequiv a+b mod p$


• $Acdot Bequiv acdot b mod p$

share|cite|improve this answer

answered Jul 24 at 6:36

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
  – Charles Quinn
  Jul 25 at 4:25
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Yes in general using congruences all algebraic rules hold for for the sum and product and we can substitute them into an expression, notably for

• $Aequiv a mod p$


• $Bequiv b mod p$

we have that

• $A+Bequiv a+b mod p$


• $Acdot Bequiv acdot b mod p$

share|cite|improve this answer

answered Jul 24 at 6:36

gimusi

65.2k73583

Yes in general using congruences all algebraic rules hold for for the sum and product and we can substitute them into an expression, notably for

• $Aequiv a mod p$


• $Bequiv b mod p$

we have that

• $A+Bequiv a+b mod p$


• $Acdot Bequiv acdot b mod p$

share|cite|improve this answer

answered Jul 24 at 6:36

gimusi

65.2k73583

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 6:36

gimusi

65.2k73583

answered Jul 24 at 6:36

gimusi

65.2k73583

answered Jul 24 at 6:36

gimusi

65.2k73583

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
  – Charles Quinn
  Jul 25 at 4:25
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
  – Charles Quinn
  Jul 25 at 4:25
  
  

  
  
  
  

Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
– Charles Quinn
Jul 25 at 4:25

Thank you. I do remember in foundations that you can manipulate congruences like that. I wasn't sure. Thanks again.
– Charles Quinn
Jul 25 at 4:25

add a comment | 

up vote
0
down vote

No. It would be very bad notation in some cases.

For example, if you replace $a^p$ by $apmod p$ in the equation $$a^ptimes5=5a^p$$ then you get $$apmod ptimes5=5apmod p$$ so the RHS would be very ambiguous. Do you mean $5apmod p$ or $5times[apmod p]$?

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I do, then I must be careful.
  – Charles Quinn
  Jul 24 at 22:32
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

No. It would be very bad notation in some cases.

For example, if you replace $a^p$ by $apmod p$ in the equation $$a^ptimes5=5a^p$$ then you get $$apmod ptimes5=5apmod p$$ so the RHS would be very ambiguous. Do you mean $5apmod p$ or $5times[apmod p]$?

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I do, then I must be careful.
  – Charles Quinn
  Jul 24 at 22:32
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

No. It would be very bad notation in some cases.

For example, if you replace $a^p$ by $apmod p$ in the equation $$a^ptimes5=5a^p$$ then you get $$apmod ptimes5=5apmod p$$ so the RHS would be very ambiguous. Do you mean $5apmod p$ or $5times[apmod p]$?

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

No. It would be very bad notation in some cases.

For example, if you replace $a^p$ by $apmod p$ in the equation $$a^ptimes5=5a^p$$ then you get $$apmod ptimes5=5apmod p$$ so the RHS would be very ambiguous. Do you mean $5apmod p$ or $5times[apmod p]$?

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

answered Jul 24 at 6:39

TheSimpliFire

9,61461851

9,61461851

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I do, then I must be careful.
  – Charles Quinn
  Jul 24 at 22:32
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I do, then I must be careful.
  – Charles Quinn
  Jul 24 at 22:32
  
  

  
  
  
  

If I do, then I must be careful.
– Charles Quinn
Jul 24 at 22:32

If I do, then I must be careful.
– Charles Quinn
Jul 24 at 22:32

add a comment | 

up vote
0
down vote

So, what about taking the Pythagorean triples as an example. We know for Pythagorean triples $x,y,$ and $z$ are integers in $x^2+y^2=z^2$. By Fermats little theorem, we know

• $a^2equiva (textmod 2)$ where $ainmathbbz$

So, $x^2+y^2=z^2$ yields,

• $x+yequivz (textmod 2)$

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

add a comment | 

up vote
0
down vote

So, what about taking the Pythagorean triples as an example. We know for Pythagorean triples $x,y,$ and $z$ are integers in $x^2+y^2=z^2$. By Fermats little theorem, we know

• $a^2equiva (textmod 2)$ where $ainmathbbz$

So, $x^2+y^2=z^2$ yields,

• $x+yequivz (textmod 2)$

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

add a comment | 

up vote
0
down vote

up vote
0
down vote

So, what about taking the Pythagorean triples as an example. We know for Pythagorean triples $x,y,$ and $z$ are integers in $x^2+y^2=z^2$. By Fermats little theorem, we know

• $a^2equiva (textmod 2)$ where $ainmathbbz$

So, $x^2+y^2=z^2$ yields,

• $x+yequivz (textmod 2)$

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

So, what about taking the Pythagorean triples as an example. We know for Pythagorean triples $x,y,$ and $z$ are integers in $x^2+y^2=z^2$. By Fermats little theorem, we know

• $a^2equiva (textmod 2)$ where $ainmathbbz$

So, $x^2+y^2=z^2$ yields,

• $x+yequivz (textmod 2)$

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 23:08

Charles Quinn

533

answered Jul 24 at 23:08

Charles Quinn

533

answered Jul 24 at 23:08

Charles Quinn

533

533

add a comment | 

add a comment | 

 

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