Mixing
Distinct prime number solution of an equation
$begingroup$
I am trying to solve the following question:
$$textrm{Find all distinct prime numbers } p, q textrm{ and } r textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 equiv p^2 equiv 1(p geq 5), 3(p = 3), 4(p = 2) mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3times(1, 3)-5times(1, 3)-4times(1, 3, 4) equiv 2 mod{6}$$
Possible solutions would be
$$begin{align}
3times1-5times3-4times1&equiv\
3times1-5times3-4times4&equiv\
3times3-5times3-4times1&equiv\
3times3-5times3-4times4&equiv2mod{6}
end{align}$$
So $q equiv 3 mod{6} Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
elementary-number-theory
edited Jan 19 at 12:44
mrtaurho
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following question:
$$textrm{Find all distinct prime numbers } p, q textrm{ and } r textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 equiv p^2 equiv 1(p geq 5), 3(p = 3), 4(p = 2) mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3times(1, 3)-5times(1, 3)-4times(1, 3, 4) equiv 2 mod{6}$$
Possible solutions would be
$$begin{align}
3times1-5times3-4times1&equiv\
3times1-5times3-4times4&equiv\
3times3-5times3-4times1&equiv\
3times3-5times3-4times4&equiv2mod{6}
end{align}$$
So $q equiv 3 mod{6} Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
elementary-number-theory
edited Jan 19 at 12:44
mrtaurho
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
$endgroup$
2
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49
add a comment |
$begingroup$
I am trying to solve the following question:
$$textrm{Find all distinct prime numbers } p, q textrm{ and } r textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 equiv p^2 equiv 1(p geq 5), 3(p = 3), 4(p = 2) mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3times(1, 3)-5times(1, 3)-4times(1, 3, 4) equiv 2 mod{6}$$
Possible solutions would be
$$begin{align}
3times1-5times3-4times1&equiv\
3times1-5times3-4times4&equiv\
3times3-5times3-4times1&equiv\
3times3-5times3-4times4&equiv2mod{6}
end{align}$$
So $q equiv 3 mod{6} Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
elementary-number-theory
edited Jan 19 at 12:44
mrtaurho
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
$endgroup$
I am trying to solve the following question:
$$textrm{Find all distinct prime numbers } p, q textrm{ and } r textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 equiv p^2 equiv 1(p geq 5), 3(p = 3), 4(p = 2) mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3times(1, 3)-5times(1, 3)-4times(1, 3, 4) equiv 2 mod{6}$$
Possible solutions would be
$$begin{align}
3times1-5times3-4times1&equiv\
3times1-5times3-4times4&equiv\
3times3-5times3-4times1&equiv\
3times3-5times3-4times4&equiv2mod{6}
end{align}$$
So $q equiv 3 mod{6} Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p geq 5, q = 3, r geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
elementary-number-theory
elementary-number-theory
edited Jan 19 at 12:44
mrtaurho
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
edited Jan 19 at 12:44
mrtaurho
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
edited Jan 19 at 12:44
mrtaurho
5,58051440
edited Jan 19 at 12:44
mrtaurho
5,58051440
edited Jan 19 at 12:44
mrtaurho
5,58051440
5,58051440
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
asked Jan 19 at 12:30
Gareth MaGareth Ma
526215
526215
2
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49
add a comment |
2
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49
2
2
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 equiv 1 pmod{5}$$
or $$r^2 equiv -2 pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3cdot 5^4-5cdot 3^4 -4r^2 = 26$$ to get $r=19.$
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
$endgroup$
add a comment |
$begingroup$
$$3p^4=5q^4+4r^2+26$$
If $sne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5mid 3p^4implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3cdot 5^4 -5cdot 2^4 -26= 4r^2implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 implies q=3 implies p=...$
If $r=5$ we have $3p^4=5q^4+126 implies q=3 implies p=...$
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
$endgroup$
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 equiv 1 pmod{5}$$
or $$r^2 equiv -2 pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3cdot 5^4-5cdot 3^4 -4r^2 = 26$$ to get $r=19.$
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
$endgroup$
add a comment |
$begingroup$
If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 equiv 1 pmod{5}$$
or $$r^2 equiv -2 pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3cdot 5^4-5cdot 3^4 -4r^2 = 26$$ to get $r=19.$
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
$endgroup$
add a comment |
$begingroup$
If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 equiv 1 pmod{5}$$
or $$r^2 equiv -2 pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3cdot 5^4-5cdot 3^4 -4r^2 = 26$$ to get $r=19.$
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
$endgroup$
If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 equiv 1 pmod{5}$$
or $$r^2 equiv -2 pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solve
$$3cdot 5^4-5cdot 3^4 -4r^2 = 26$$ to get $r=19.$
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
answered Jan 19 at 15:11
B. GoddardB. Goddard
19.3k21441
19.3k21441
add a comment |
add a comment |
$begingroup$
$$3p^4=5q^4+4r^2+26$$
If $sne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5mid 3p^4implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3cdot 5^4 -5cdot 2^4 -26= 4r^2implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 implies q=3 implies p=...$
If $r=5$ we have $3p^4=5q^4+126 implies q=3 implies p=...$
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
$endgroup$
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
add a comment |
$begingroup$
$$3p^4=5q^4+4r^2+26$$
If $sne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5mid 3p^4implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3cdot 5^4 -5cdot 2^4 -26= 4r^2implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 implies q=3 implies p=...$
If $r=5$ we have $3p^4=5q^4+126 implies q=3 implies p=...$
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
$endgroup$
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
add a comment |
$begingroup$
$$3p^4=5q^4+4r^2+26$$
If $sne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5mid 3p^4implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3cdot 5^4 -5cdot 2^4 -26= 4r^2implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 implies q=3 implies p=...$
If $r=5$ we have $3p^4=5q^4+126 implies q=3 implies p=...$
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
$endgroup$
$$3p^4=5q^4+4r^2+26$$
If $sne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5mid 3p^4implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3cdot 5^4 -5cdot 2^4 -26= 4r^2implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 implies q=3 implies p=...$
If $r=5$ we have $3p^4=5q^4+126 implies q=3 implies p=...$
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
answered Jan 19 at 15:47
greedoidgreedoid
45k1157112
45k1157112
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
add a comment |
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
$begingroup$
"This can be only if q=2" I think it is the exact opposite?
$endgroup$
– Gareth Ma
Jan 20 at 1:53
add a comment |
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2
$begingroup$
Try reducing the equation modulo $5$.
$endgroup$
– B. Goddard
Jan 19 at 12:53
$begingroup$
Do you want to answer it so you get the correct answer? Thanks a lot <3
$endgroup$
– Gareth Ma
Jan 19 at 14:49