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If $ab mid c(c^2-c+1)$ and $c^2+1 mid a+b$ then prove that $a, b=c, c^2-c+1 $
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If $ab mid c(c^2-c+1)$ and $c^2+1 mid a+b$ then prove that $a, b=c, c^2-c+1 $ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \ c(a+b)=dc(c^2-c+1)+dc^2 \ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
number-theory elementary-number-theory contest-math diophantine-equations
asked Jul 23 at 11:02
Dedaha
1007
 |Â
show 4 more comments
up vote
8
down vote
favorite
If $ab mid c(c^2-c+1)$ and $c^2+1 mid a+b$ then prove that $a, b=c, c^2-c+1 $ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \ c(a+b)=dc(c^2-c+1)+dc^2 \ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
number-theory elementary-number-theory contest-math diophantine-equations
asked Jul 23 at 11:02
Dedaha
1007
Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
1
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
1
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
2
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
2
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37
 |Â
show 4 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
If $ab mid c(c^2-c+1)$ and $c^2+1 mid a+b$ then prove that $a, b=c, c^2-c+1 $ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \ c(a+b)=dc(c^2-c+1)+dc^2 \ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
number-theory elementary-number-theory contest-math diophantine-equations
asked Jul 23 at 11:02
Dedaha
1007
If $ab mid c(c^2-c+1)$ and $c^2+1 mid a+b$ then prove that $a, b=c, c^2-c+1 $ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \ c(a+b)=dc(c^2-c+1)+dc^2 \ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
number-theory elementary-number-theory contest-math diophantine-equations
asked Jul 23 at 11:02
Dedaha
1007
edited Jul 23 at 18:29
asked Jul 23 at 11:02
Dedaha
1007
asked Jul 23 at 11:02
Dedaha
1007
asked Jul 23 at 11:02
Dedaha
1007
1007
Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
1
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
1
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
2
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
2
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37
 |Â
show 4 more comments
Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
1
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
1
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
2
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
2
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37
Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
1
1
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
1
1
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
2
2
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
2
2
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
HINT.-For some positive integers $m,n$ we have
$$begincasesa+b=m(c^2+1)\abn=c(c^2-c+1)endcasesRightarrow c^2-fraca+bmc+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $cina,b$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+dfraca+bmX+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
answered Jul 23 at 21:46
Piquito
17.4k31234
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
add a comment |Â
up vote
2
down vote
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=frac2mc+(c^2-c+1)/22m^2(c^2+1)^2$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
answered Jul 25 at 17:59
sirous dehmami
294
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
HINT.-For some positive integers $m,n$ we have
$$begincasesa+b=m(c^2+1)\abn=c(c^2-c+1)endcasesRightarrow c^2-fraca+bmc+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $cina,b$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+dfraca+bmX+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
answered Jul 23 at 21:46
Piquito
17.4k31234
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
add a comment |Â
up vote
2
down vote
accepted
HINT.-For some positive integers $m,n$ we have
$$begincasesa+b=m(c^2+1)\abn=c(c^2-c+1)endcasesRightarrow c^2-fraca+bmc+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $cina,b$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+dfraca+bmX+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
answered Jul 23 at 21:46
Piquito
17.4k31234
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT.-For some positive integers $m,n$ we have
$$begincasesa+b=m(c^2+1)\abn=c(c^2-c+1)endcasesRightarrow c^2-fraca+bmc+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $cina,b$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+dfraca+bmX+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
answered Jul 23 at 21:46
Piquito
17.4k31234
HINT.-For some positive integers $m,n$ we have
$$begincasesa+b=m(c^2+1)\abn=c(c^2-c+1)endcasesRightarrow c^2-fraca+bmc+abn=0$$
If $m$ and $n$ are equal to $1$ then $c$ is root of the equation $X^2-(a+b)X+ab=0$ from which clearly $cina,b$. It easily follows that $a$ or $b$ is equal to $c^2-c+1$.
If both $m$ and $n$ is greater than $1$ then $c$ is root of $X^2+dfraca+bmX+abn=0$ where the sum of the two roots shrinks while the product is enlarged. This is not possible for positive integers.
I leave for the O.P. the case just one of the $m,n$ is equal to $1$
Finally the proof is clear (it has been done already above) because necessarily the positive integers $m$ and $n$ are equal to $1$.
answered Jul 23 at 21:46
Piquito
17.4k31234
answered Jul 23 at 21:46
Piquito
17.4k31234
answered Jul 23 at 21:46
Piquito
17.4k31234
answered Jul 23 at 21:46
Piquito
17.4k31234
17.4k31234
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
add a comment |Â
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
1
1
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
If $m,n>1$, then $c$ is a root of $x^2-[(a+b)/m]x+abn=0$, but the other root can be a non-integer! So using sum and product of roots gives no information about $c$.
– Dedaha
Jul 24 at 10:19
1
1
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
Can you give an example of such a situation? With the two restrictions given, of course.
– Piquito
Jul 24 at 13:57
1
1
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
If one root of a monic quadratic with integer coefficients is an integer, the other root must be an integer too (since the sum of both roots is one of the integer coefficients).
– Sameer Kailasa
Jul 24 at 14:28
add a comment |Â
up vote
2
down vote
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=frac2mc+(c^2-c+1)/22m^2(c^2+1)^2$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
answered Jul 25 at 17:59
sirous dehmami
294
add a comment |Â
up vote
2
down vote
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=frac2mc+(c^2-c+1)/22m^2(c^2+1)^2$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
answered Jul 25 at 17:59
sirous dehmami
294
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=frac2mc+(c^2-c+1)/22m^2(c^2+1)^2$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
answered Jul 25 at 17:59
sirous dehmami
294
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=frac2mc+(c^2-c+1)/22m^2(c^2+1)^2$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
answered Jul 25 at 17:59
sirous dehmami
294
answered Jul 25 at 17:59
sirous dehmami
294
answered Jul 25 at 17:59
sirous dehmami
294
answered Jul 25 at 17:59
sirous dehmami
294
294
add a comment |Â
add a comment |Â
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Is it an ongoing contest ?
– Peter
Jul 23 at 11:06
1
No, no at all.. Its from a book, I only saw the picture of this question. A friend sent it in a math group.
– Dedaha
Jul 23 at 11:09
1
This is just an initial reaction, but it looks like a Viete jumping problem (this may have been what you were thinking by taking the discriminant, though).
– Rellek
Jul 23 at 11:13
2
Can you clarify your question? When you write $(a,b)$ do you mean the ordered pair or are you referring to the gcd? If the latter, then it is very odd as $gcd(c,c^2-c+1)$ is obviously $1$. If the former then your claim is false...if nothing else we might have $(a,b)=(c^2-c+1,c)$.
– lulu
Jul 23 at 11:30
2
What might help : $$x^2+(c^2+1)x+c(c^2-c+1)=(x+c)(x+c^2-c+1)$$
– Peter
Jul 23 at 12:37