Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$.
$begingroup$
Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
number-theory prime-numbers modular-arithmetic contest-math divisibility
asked Jan 5 at 1:29
user574848user574848
689118
$endgroup$
add a comment |
$begingroup$
Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
number-theory prime-numbers modular-arithmetic contest-math divisibility
asked Jan 5 at 1:29
user574848user574848
689118
$endgroup$
2
$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43
$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58
$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43
add a comment |
$begingroup$
Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
number-theory prime-numbers modular-arithmetic contest-math divisibility
asked Jan 5 at 1:29
user574848user574848
689118
$endgroup$
Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
number-theory prime-numbers modular-arithmetic contest-math divisibility
number-theory prime-numbers modular-arithmetic contest-math divisibility
asked Jan 5 at 1:29
user574848user574848
689118
asked Jan 5 at 1:29
user574848user574848
689118
edited Jan 5 at 2:37
user574848
asked Jan 5 at 1:29
user574848user574848
689118
asked Jan 5 at 1:29
user574848user574848
689118
asked Jan 5 at 1:29
user574848user574848
689118
689118
2
$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43
$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58
$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43
add a comment |
2
$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43
$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58
$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43
2
2
$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43
$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43
$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58
$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58
$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43
$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$3a - 1equiv bpmod p$
for any prime $p ne 3$
$3$ has a multiplicative inverse.
$a equiv 3^{-1}b + 3^{-1}pmod p$
And if $p=3$ let $a = 1$
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
add a comment |
$begingroup$
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2equiv-99pmod p$
$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$
$iff2b-1equivpm3(2a-1)$
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
add a comment |
$begingroup$
Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.
answered Jan 11 at 15:14
AaronAaron
1,992415
$endgroup$
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
$begingroup$
$3a - 1equiv bpmod p$
for any prime $p ne 3$
$3$ has a multiplicative inverse.
$a equiv 3^{-1}b + 3^{-1}pmod p$
And if $p=3$ let $a = 1$
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
add a comment |
$begingroup$
$3a - 1equiv bpmod p$
for any prime $p ne 3$
$3$ has a multiplicative inverse.
$a equiv 3^{-1}b + 3^{-1}pmod p$
And if $p=3$ let $a = 1$
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
add a comment |
$begingroup$
$3a - 1equiv bpmod p$
for any prime $p ne 3$
$3$ has a multiplicative inverse.
$a equiv 3^{-1}b + 3^{-1}pmod p$
And if $p=3$ let $a = 1$
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
$endgroup$
$3a - 1equiv bpmod p$
for any prime $p ne 3$
$3$ has a multiplicative inverse.
$a equiv 3^{-1}b + 3^{-1}pmod p$
And if $p=3$ let $a = 1$
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
answered Jan 5 at 2:10
Doug MDoug M
45.3k31954
45.3k31954
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
add a comment |
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
What's a multiplicative inverse?
$endgroup$
– user574848
Jan 5 at 2:50
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
$begingroup$
The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
$endgroup$
– Doug M
Jan 5 at 3:34
add a comment |
$begingroup$
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2equiv-99pmod p$
$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$
$iff2b-1equivpm3(2a-1)$
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
add a comment |
$begingroup$
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2equiv-99pmod p$
$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$
$iff2b-1equivpm3(2a-1)$
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
add a comment |
$begingroup$
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2equiv-99pmod p$
$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$
$iff2b-1equivpm3(2a-1)$
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2equiv-99pmod p$
$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$
$iff2b-1equivpm3(2a-1)$
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
answered Jan 5 at 2:28
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
add a comment |
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
$begingroup$
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:00
add a comment |
$begingroup$
Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.
answered Jan 11 at 15:14
AaronAaron
1,992415
$endgroup$
add a comment |
$begingroup$
Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.
answered Jan 11 at 15:14
AaronAaron
1,992415
$endgroup$
add a comment |
$begingroup$
Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.
answered Jan 11 at 15:14
AaronAaron
1,992415
$endgroup$
Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.
answered Jan 11 at 15:14
AaronAaron
1,992415
answered Jan 11 at 15:14
AaronAaron
1,992415
answered Jan 11 at 15:14
AaronAaron
1,992415
answered Jan 11 at 15:14
AaronAaron
1,992415
1,992415
add a comment |
add a comment |
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2
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You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
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– Doug M
Jan 5 at 1:43
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@DougM thanks, fixed
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– user574848
Jan 5 at 1:58
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What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
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– Will Jagy
Jan 5 at 2:43