$a^2+bc \equiv a (mod 37)$
$b(a+d) \equiv b (mod 37)$
$c(a+d) \equiv c (mod 37)$
$bc+d^2 \equiv d (mod 37)$
$ad-bc \equiv 1 (mod 37)$
Shitty Number theory
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Souvik saha
- Posts:6
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Find the total number of solutions to the following system of equations:
$a^2+bc \equiv a (mod 37)$
$b(a+d) \equiv b (mod 37)$
$c(a+d) \equiv c (mod 37)$
$bc+d^2 \equiv d (mod 37)$
$ad-bc \equiv 1 (mod 37)$
$a^2+bc \equiv a (mod 37)$
$b(a+d) \equiv b (mod 37)$
$c(a+d) \equiv c (mod 37)$
$bc+d^2 \equiv d (mod 37)$
$ad-bc \equiv 1 (mod 37)$
If you can’t make it good, at least make it look good . – Bill Gates
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Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Shitty Number theory
From $(2)$ we get $37|b(a+d-1)$. So $37|b$ or $37|a+d-1$.
If $37|a+d-1$ we get $ad-bc \equiv ad-(a-a^2) \equiv a(a+d-1) \equiv 0 \pmod{37}$, a contradiction so $37|b$.
Similarly from $(3)$ we get $37|c$.
So $(1),(4),(5)$ can be reduced to $37|a(a-1),37|d(d-1),ad \equiv 1\pmod{37}$.
From the third expression $37 \nmid a,d$. So $37|a-1,37|d-1$.
Hence $(a,b,c,d) \equiv 1,0,0,1 \pmod{37}$.
It's easy to check that these indeed statisfy the given congruences.
Note: This system works for any prime.
If $37|a+d-1$ we get $ad-bc \equiv ad-(a-a^2) \equiv a(a+d-1) \equiv 0 \pmod{37}$, a contradiction so $37|b$.
Similarly from $(3)$ we get $37|c$.
So $(1),(4),(5)$ can be reduced to $37|a(a-1),37|d(d-1),ad \equiv 1\pmod{37}$.
From the third expression $37 \nmid a,d$. So $37|a-1,37|d-1$.
Hence $(a,b,c,d) \equiv 1,0,0,1 \pmod{37}$.
It's easy to check that these indeed statisfy the given congruences.
Note: This system works for any prime.
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