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Can $ab+cd$ be an integer if $a, b, c, d$ are non integers and each 3 sum to an integer [closed]

General set of integer solutions $(p,q)$ to $1 = pa + qb$ for integers $a,b$ such that $gcd(a,b)=1$Is every non-square integer a primitive root modulo some odd prime?Show that if a and b are positive integers, then $(a + frac12)^n + (b + frac12)^n$ is an integer for only finitely many positive integers n.How do I prove that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite?Prove that there exist $n$ consecutive positive integers with the $i$th integer divisible by the $i$th primeShow that any non-negative rational integer may be written as the sum of two squares if it is of the form…Number of ways to express an integer N as sum of 4 integersum of all positive integers $k$ with $1 le k < n$ and $gcd(k,n)=1$How to write a given integer as sum of other integers?Find all values of $p-q$ if $p, q$ are prime and $q+1over q+pover p+1=2nover n+2$ where $n$ is a positive integer.

-2

$begingroup$

Can $ab+cd$ be an integer if $a, b, c, d$ are non integers and each 3 sum to an integer.

This question in my opinion is tricky and I am unable to do much in it. However, after trying to substitute I got that it is impossible. Any help would be appreciated

elementary-number-theory divisibility

share|cite|improve this question

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

$endgroup$

closed as off-topic by John Douma, Cesareo, Javi, Lee David Chung Lin, Eevee Trainer Mar 24 at 4:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, Cesareo, Javi, Lee David Chung Lin, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? Please add your works on the problem in the body to avoid further downvotes.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user587054: Can you please explain what have you tried so far?
  $endgroup$
  – Ertxiem
  Mar 22 at 14:39
  

  
  
  
  

add a comment | 

-2

$begingroup$

Can $ab+cd$ be an integer if $a, b, c, d$ are non integers and each 3 sum to an integer.

This question in my opinion is tricky and I am unable to do much in it. However, after trying to substitute I got that it is impossible. Any help would be appreciated

elementary-number-theory divisibility

share|cite|improve this question

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

$endgroup$

closed as off-topic by John Douma, Cesareo, Javi, Lee David Chung Lin, Eevee Trainer Mar 24 at 4:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, Cesareo, Javi, Lee David Chung Lin, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? Please add your works on the problem in the body to avoid further downvotes.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user587054: Can you please explain what have you tried so far?
  $endgroup$
  – Ertxiem
  Mar 22 at 14:39
  

  
  
  
  

add a comment | 

$begingroup$

Can $ab+cd$ be an integer if $a, b, c, d$ are non integers and each 3 sum to an integer.

This question in my opinion is tricky and I am unable to do much in it. However, after trying to substitute I got that it is impossible. Any help would be appreciated

elementary-number-theory divisibility

share|cite|improve this question

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

$endgroup$

share|cite|improve this question

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

share|cite|improve this question

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

edited Mar 22 at 15:04

Maria Mazur

50k1361125

edited Mar 22 at 15:04

Maria Mazur

50k1361125

asked Mar 22 at 14:12

user587054user587054

59411

asked Mar 22 at 14:12

user587054user587054

59411

3 Answers
3

active

oldest

votes

2

$begingroup$

It isn't possible.

Let $a$ be the least of the integers (possibly not unique).

We note that $$a+b+cin mathbb Z quad &quad a+b+din mathbb Zimplies c-din mathbb Z$$

Similarly the difference between any two of the variables must be an integer.

We let $b=a+n_1,,c=a+n_2,d=a+n_3$ with $n_iin mathbb Z$.

Since $a+b+cin mathbb Z$ we note that $3a+n_1+n_2in mathbb Zimplies 3ain mathbb Z$. Thus there is an integer $m$ with $a=frac m3$. Of course $mneq 0 pmod 3$, hence $mequiv pm 1 pmod 3$.

We now look at the requirement that $ab+cdin mathbb Z$. This quickly tells us that $$2a^2+a(n_1+n_2+n_3)+n_2n_3in mathbb Z$$

Thus $$2a^2+a(n_1+n_2+n_3)in mathbb Z$$

Let $n=n_1+n_2+n_3$. If $mequiv 1 pmod 3$ then $m=3k+1$ for some integer $k$, but this would mean that $$frac 2(9k^2+6k+1)+(9k+3)n9in Z$$ which is clearly impossible (the numerator isn't even divisible by $3$).

The case $m=3k-1$ is similar.

share|cite|improve this answer

answered Mar 22 at 14:48

lulululu

43.6k25081

$endgroup$

add a comment | 

2

$begingroup$

No. From $a+b+c=$integer and $a+b+d=$ integer we deduce that $c-d=$ integer and similarly for all the others. This says the fractional parts of $a,b,c,d$ are equal. It can only be $0,frac 13,$ or $frac 23$. The question prohibits $0$. Let $a'=lfloor a rfloor$ and similarly for the others. Then choosing the $frac 13$ case we have
$$ab+cd=left(a'+frac 13right)left(b'+frac 13right)+left(c'+frac 13right)left(d'+frac 13right)\
=a'b'+c'd'+frac 13(a'+b')+frac 19+c'd'+frac 13(c'+d')+frac 19$$
As all the primed terms are integers, there is no way to get rid of the $frac 29$. The same happens in the $frac 23$ case except the fraction is $frac 89$

share|cite|improve this answer

edited Mar 22 at 14:47

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why @student? Every $3$ sum is given to be an integer.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! I see. Sorry for making misleading comment @Ross Milikan.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:54
  
  

  
  
  
  

add a comment | 

0

$begingroup$

$$a+b+c=x$$
$$a+b+d=y$$
$$a+c+d=z$$
$$b+c+d=t$$

Then $$a+b+c+d=x+y+z+tover 3$$

So $$ a=x+y+z-2tover 3implies a= a'over 3$$

where $a'$ is an integer not divisible by $3$. Since $a-b$ is an integer we have $3mid a'-b'$ and the same for all other pairs, so we have $$a'equiv_3 b'equiv_3 c'equiv_3 d'=r$$

Since $$ab+cd= a'b'+d'c'over 9$$ so if $ab+cdin mathbbZ$ we have $9mid a'b'+c'd' $ Now we have to check folowing 2 cases:

1. case $r=1$, then $$a'b'+c'd' = (3m+1)(3n+1)+(3k+1)(3l+1) = 3(...)+2$$ which is impossibile and




2. case $r=2$, then $$a'b'+c'd' = (3m+2)(3n+2)+(3k+2)(3l+2) = 3(...)+8$$ which is again impossibile.

So such a numbers don't exists.

share|cite|improve this answer

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

2

$begingroup$

It isn't possible.

Let $a$ be the least of the integers (possibly not unique).

We note that $$a+b+cin mathbb Z quad &quad a+b+din mathbb Zimplies c-din mathbb Z$$

Similarly the difference between any two of the variables must be an integer.

We let $b=a+n_1,,c=a+n_2,d=a+n_3$ with $n_iin mathbb Z$.

Since $a+b+cin mathbb Z$ we note that $3a+n_1+n_2in mathbb Zimplies 3ain mathbb Z$. Thus there is an integer $m$ with $a=frac m3$. Of course $mneq 0 pmod 3$, hence $mequiv pm 1 pmod 3$.

We now look at the requirement that $ab+cdin mathbb Z$. This quickly tells us that $$2a^2+a(n_1+n_2+n_3)+n_2n_3in mathbb Z$$

Thus $$2a^2+a(n_1+n_2+n_3)in mathbb Z$$

Let $n=n_1+n_2+n_3$. If $mequiv 1 pmod 3$ then $m=3k+1$ for some integer $k$, but this would mean that $$frac 2(9k^2+6k+1)+(9k+3)n9in Z$$ which is clearly impossible (the numerator isn't even divisible by $3$).

The case $m=3k-1$ is similar.

share|cite|improve this answer

answered Mar 22 at 14:48

lulululu

43.6k25081

$endgroup$

add a comment | 

2

$begingroup$

It isn't possible.

Let $a$ be the least of the integers (possibly not unique).

We note that $$a+b+cin mathbb Z quad &quad a+b+din mathbb Zimplies c-din mathbb Z$$

Similarly the difference between any two of the variables must be an integer.

We let $b=a+n_1,,c=a+n_2,d=a+n_3$ with $n_iin mathbb Z$.

Since $a+b+cin mathbb Z$ we note that $3a+n_1+n_2in mathbb Zimplies 3ain mathbb Z$. Thus there is an integer $m$ with $a=frac m3$. Of course $mneq 0 pmod 3$, hence $mequiv pm 1 pmod 3$.

We now look at the requirement that $ab+cdin mathbb Z$. This quickly tells us that $$2a^2+a(n_1+n_2+n_3)+n_2n_3in mathbb Z$$

Thus $$2a^2+a(n_1+n_2+n_3)in mathbb Z$$

Let $n=n_1+n_2+n_3$. If $mequiv 1 pmod 3$ then $m=3k+1$ for some integer $k$, but this would mean that $$frac 2(9k^2+6k+1)+(9k+3)n9in Z$$ which is clearly impossible (the numerator isn't even divisible by $3$).

The case $m=3k-1$ is similar.

share|cite|improve this answer

answered Mar 22 at 14:48

lulululu

43.6k25081

$endgroup$

add a comment | 

$begingroup$

It isn't possible.

Let $a$ be the least of the integers (possibly not unique).

We note that $$a+b+cin mathbb Z quad &quad a+b+din mathbb Zimplies c-din mathbb Z$$

Similarly the difference between any two of the variables must be an integer.

We let $b=a+n_1,,c=a+n_2,d=a+n_3$ with $n_iin mathbb Z$.

Since $a+b+cin mathbb Z$ we note that $3a+n_1+n_2in mathbb Zimplies 3ain mathbb Z$. Thus there is an integer $m$ with $a=frac m3$. Of course $mneq 0 pmod 3$, hence $mequiv pm 1 pmod 3$.

We now look at the requirement that $ab+cdin mathbb Z$. This quickly tells us that $$2a^2+a(n_1+n_2+n_3)+n_2n_3in mathbb Z$$

Thus $$2a^2+a(n_1+n_2+n_3)in mathbb Z$$

Let $n=n_1+n_2+n_3$. If $mequiv 1 pmod 3$ then $m=3k+1$ for some integer $k$, but this would mean that $$frac 2(9k^2+6k+1)+(9k+3)n9in Z$$ which is clearly impossible (the numerator isn't even divisible by $3$).

The case $m=3k-1$ is similar.

share|cite|improve this answer

answered Mar 22 at 14:48

lulululu

43.6k25081

$endgroup$

share|cite|improve this answer

answered Mar 22 at 14:48

lulululu

43.6k25081

answered Mar 22 at 14:48

lulululu

43.6k25081

answered Mar 22 at 14:48

lulululu

43.6k25081

2

$begingroup$

No. From $a+b+c=$integer and $a+b+d=$ integer we deduce that $c-d=$ integer and similarly for all the others. This says the fractional parts of $a,b,c,d$ are equal. It can only be $0,frac 13,$ or $frac 23$. The question prohibits $0$. Let $a'=lfloor a rfloor$ and similarly for the others. Then choosing the $frac 13$ case we have
$$ab+cd=left(a'+frac 13right)left(b'+frac 13right)+left(c'+frac 13right)left(d'+frac 13right)\
=a'b'+c'd'+frac 13(a'+b')+frac 19+c'd'+frac 13(c'+d')+frac 19$$
As all the primed terms are integers, there is no way to get rid of the $frac 29$. The same happens in the $frac 23$ case except the fraction is $frac 89$

share|cite|improve this answer

edited Mar 22 at 14:47

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why @student? Every $3$ sum is given to be an integer.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! I see. Sorry for making misleading comment @Ross Milikan.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:54
  
  

  
  
  
  

add a comment | 

2

$begingroup$

No. From $a+b+c=$integer and $a+b+d=$ integer we deduce that $c-d=$ integer and similarly for all the others. This says the fractional parts of $a,b,c,d$ are equal. It can only be $0,frac 13,$ or $frac 23$. The question prohibits $0$. Let $a'=lfloor a rfloor$ and similarly for the others. Then choosing the $frac 13$ case we have
$$ab+cd=left(a'+frac 13right)left(b'+frac 13right)+left(c'+frac 13right)left(d'+frac 13right)\
=a'b'+c'd'+frac 13(a'+b')+frac 19+c'd'+frac 13(c'+d')+frac 19$$
As all the primed terms are integers, there is no way to get rid of the $frac 29$. The same happens in the $frac 23$ case except the fraction is $frac 89$

share|cite|improve this answer

edited Mar 22 at 14:47

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why @student? Every $3$ sum is given to be an integer.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! I see. Sorry for making misleading comment @Ross Milikan.
  $endgroup$
  – Dbchatto67
  Mar 22 at 14:54
  
  

  
  
  
  

add a comment | 

$begingroup$

No. From $a+b+c=$integer and $a+b+d=$ integer we deduce that $c-d=$ integer and similarly for all the others. This says the fractional parts of $a,b,c,d$ are equal. It can only be $0,frac 13,$ or $frac 23$. The question prohibits $0$. Let $a'=lfloor a rfloor$ and similarly for the others. Then choosing the $frac 13$ case we have
$$ab+cd=left(a'+frac 13right)left(b'+frac 13right)+left(c'+frac 13right)left(d'+frac 13right)\
=a'b'+c'd'+frac 13(a'+b')+frac 19+c'd'+frac 13(c'+d')+frac 19$$
As all the primed terms are integers, there is no way to get rid of the $frac 29$. The same happens in the $frac 23$ case except the fraction is $frac 89$

share|cite|improve this answer

edited Mar 22 at 14:47

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

$endgroup$

share|cite|improve this answer

edited Mar 22 at 14:47

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

answered Mar 22 at 14:20

Ross MillikanRoss Millikan

301k24200375

0

$begingroup$

$$a+b+c=x$$
$$a+b+d=y$$
$$a+c+d=z$$
$$b+c+d=t$$

Then $$a+b+c+d=x+y+z+tover 3$$

So $$ a=x+y+z-2tover 3implies a= a'over 3$$

where $a'$ is an integer not divisible by $3$. Since $a-b$ is an integer we have $3mid a'-b'$ and the same for all other pairs, so we have $$a'equiv_3 b'equiv_3 c'equiv_3 d'=r$$

Since $$ab+cd= a'b'+d'c'over 9$$ so if $ab+cdin mathbbZ$ we have $9mid a'b'+c'd' $ Now we have to check folowing 2 cases:

1. case $r=1$, then $$a'b'+c'd' = (3m+1)(3n+1)+(3k+1)(3l+1) = 3(...)+2$$ which is impossibile and




2. case $r=2$, then $$a'b'+c'd' = (3m+2)(3n+2)+(3k+2)(3l+2) = 3(...)+8$$ which is again impossibile.

So such a numbers don't exists.

share|cite|improve this answer

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

0

$begingroup$

$$a+b+c=x$$
$$a+b+d=y$$
$$a+c+d=z$$
$$b+c+d=t$$

Then $$a+b+c+d=x+y+z+tover 3$$

So $$ a=x+y+z-2tover 3implies a= a'over 3$$

where $a'$ is an integer not divisible by $3$. Since $a-b$ is an integer we have $3mid a'-b'$ and the same for all other pairs, so we have $$a'equiv_3 b'equiv_3 c'equiv_3 d'=r$$

Since $$ab+cd= a'b'+d'c'over 9$$ so if $ab+cdin mathbbZ$ we have $9mid a'b'+c'd' $ Now we have to check folowing 2 cases:

1. case $r=1$, then $$a'b'+c'd' = (3m+1)(3n+1)+(3k+1)(3l+1) = 3(...)+2$$ which is impossibile and




2. case $r=2$, then $$a'b'+c'd' = (3m+2)(3n+2)+(3k+2)(3l+2) = 3(...)+8$$ which is again impossibile.

So such a numbers don't exists.

share|cite|improve this answer

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

$begingroup$

$$a+b+c=x$$
$$a+b+d=y$$
$$a+c+d=z$$
$$b+c+d=t$$

Then $$a+b+c+d=x+y+z+tover 3$$

So $$ a=x+y+z-2tover 3implies a= a'over 3$$

where $a'$ is an integer not divisible by $3$. Since $a-b$ is an integer we have $3mid a'-b'$ and the same for all other pairs, so we have $$a'equiv_3 b'equiv_3 c'equiv_3 d'=r$$

Since $$ab+cd= a'b'+d'c'over 9$$ so if $ab+cdin mathbbZ$ we have $9mid a'b'+c'd' $ Now we have to check folowing 2 cases:

1. case $r=1$, then $$a'b'+c'd' = (3m+1)(3n+1)+(3k+1)(3l+1) = 3(...)+2$$ which is impossibile and




2. case $r=2$, then $$a'b'+c'd' = (3m+2)(3n+2)+(3k+2)(3l+2) = 3(...)+8$$ which is again impossibile.

So such a numbers don't exists.

share|cite|improve this answer

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125

answered Mar 22 at 15:03

Maria MazurMaria Mazur

50k1361125