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When is $m^2k^2(c^2+1)^2-4mc(c^2-c+1)$ a perfect square?
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Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,cle 1 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1mid a+b$ and $abmid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
number-theory elementary-number-theory square-numbers
asked Jul 25 at 15:42
Peter
45k938119
add a comment |Â
up vote
4
down vote
favorite
Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,cle 1 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1mid a+b$ and $abmid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
number-theory elementary-number-theory square-numbers
asked Jul 25 at 15:42
Peter
45k938119
See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,cle 1 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1mid a+b$ and $abmid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
number-theory elementary-number-theory square-numbers
asked Jul 25 at 15:42
Peter
45k938119
Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,cle 1 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1mid a+b$ and $abmid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
number-theory elementary-number-theory square-numbers
asked Jul 25 at 15:42
Peter
45k938119
edited Jul 25 at 19:21
asked Jul 25 at 15:42
Peter
45k938119
asked Jul 25 at 15:42
Peter
45k938119
asked Jul 25 at 15:42
Peter
45k938119
45k938119
See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38
add a comment |Â
See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38
See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38
add a comment |Â
1 Answer
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1
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If we write $d=gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = frac4c(c^2-c+1)d
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
left( t^2 d (c^2+1) right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
left( t^2 d (c^2+1) right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 geq a_4+2$. But
$$
left( t^2 d (c^2+1) right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
answered Jul 26 at 14:47
Mike Bennett
2,24468
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If we write $d=gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = frac4c(c^2-c+1)d
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
left( t^2 d (c^2+1) right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
left( t^2 d (c^2+1) right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 geq a_4+2$. But
$$
left( t^2 d (c^2+1) right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
answered Jul 26 at 14:47
Mike Bennett
2,24468
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
add a comment |Â
up vote
1
down vote
If we write $d=gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = frac4c(c^2-c+1)d
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
left( t^2 d (c^2+1) right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
left( t^2 d (c^2+1) right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 geq a_4+2$. But
$$
left( t^2 d (c^2+1) right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
answered Jul 26 at 14:47
Mike Bennett
2,24468
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If we write $d=gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = frac4c(c^2-c+1)d
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
left( t^2 d (c^2+1) right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
left( t^2 d (c^2+1) right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 geq a_4+2$. But
$$
left( t^2 d (c^2+1) right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
answered Jul 26 at 14:47
Mike Bennett
2,24468
If we write $d=gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = frac4c(c^2-c+1)d
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
left( t^2 d (c^2+1) right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
left( t^2 d (c^2+1) right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 geq a_4+2$. But
$$
left( t^2 d (c^2+1) right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
answered Jul 26 at 14:47
Mike Bennett
2,24468
answered Jul 26 at 14:47
Mike Bennett
2,24468
answered Jul 26 at 14:47
Mike Bennett
2,24468
answered Jul 26 at 14:47
Mike Bennett
2,24468
2,24468
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
add a comment |Â
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
Exellent! Thanks sir. I have a question! Why $a_3$ and $a_4$ have same parity?
– Dedaha
Jul 31 at 8:02
add a comment |Â
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See here : math.stackexchange.com/questions/2860229/…
– Peter
Jul 25 at 16:54
Anyone thinking Vieta jumping on $m$? I don't have the time to try it right now but if we can use it to prove $m=1$ then we will have made progress.
– Cataline
Jul 25 at 19:36
@Cataline Of course I tried this, but without success.
– Peter
Jul 25 at 19:38