Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$.
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 yesterday
user574848
16415
add a comment |
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 yesterday
user574848
16415
2
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
@DougM thanks, fixed
– user574848
yesterday
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 $-99$
– Will Jagy
yesterday
add a comment |
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 yesterday
user574848
16415
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 yesterday
user574848
16415
asked yesterday
user574848
16415
edited yesterday
asked yesterday
user574848
16415
asked yesterday
user574848
16415
asked yesterday
user574848
16415
16415
2
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
@DougM thanks, fixed
– user574848
yesterday
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 $-99$
– Will Jagy
yesterday
add a comment |
2
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
@DougM thanks, fixed
– user574848
yesterday
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 $-99$
– Will Jagy
yesterday
2
2
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
@DougM thanks, fixed
– user574848
yesterday
@DougM thanks, fixed
– user574848
yesterday
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 $-99$
– Will Jagy
yesterday
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 $-99$
– Will Jagy
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$3a - 1equiv bpmod p$
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 yesterday
Doug M
44.2k31854
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
add a comment |
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ 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 yesterday
lab bhattacharjee
223k15156274
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
add a comment |
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2 Answers
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$3a - 1equiv bpmod p$
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 yesterday
Doug M
44.2k31854
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
add a comment |
$3a - 1equiv bpmod p$
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 yesterday
Doug M
44.2k31854
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
add a comment |
$3a - 1equiv bpmod p$
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 yesterday
Doug M
44.2k31854
$3a - 1equiv bpmod p$
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 yesterday
Doug M
44.2k31854
answered yesterday
Doug M
44.2k31854
answered yesterday
Doug M
44.2k31854
answered yesterday
Doug M
44.2k31854
44.2k31854
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
add a comment |
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
What's a multiplicative inverse?
– user574848
yesterday
What's a multiplicative inverse?
– user574848
yesterday
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.
– Doug M
yesterday
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.
– Doug M
yesterday
add a comment |
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ 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 yesterday
lab bhattacharjee
223k15156274
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
add a comment |
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ 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 yesterday
lab bhattacharjee
223k15156274
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
add a comment |
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ 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 yesterday
lab bhattacharjee
223k15156274
$a^2-a+3equiv0pmod p$
$iff(2a-1)^2equiv-11$ 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 yesterday
lab bhattacharjee
223k15156274
answered yesterday
lab bhattacharjee
223k15156274
answered yesterday
lab bhattacharjee
223k15156274
answered yesterday
lab bhattacharjee
223k15156274
223k15156274
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
add a comment |
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
@user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
– lab bhattacharjee
yesterday
add a comment |
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2
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
– Doug M
yesterday
@DougM thanks, fixed
– user574848
yesterday
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 $-99$
– Will Jagy
yesterday