How do I combine the congruences $p=\pm 1\mod 8$ and $p=\pm 1\mod 5$ into a congruence $\mod 40$ where $p$ is an odd prime? I can see by writing out values the answer is $p=\pm 1\mod 40$ and $p=\pm 9\mod 40$
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If you can see the answer where is the question? – Dietrich Burde Mar 02 '20 at 12:47
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How do I prove that this is the answer*? – Loobear23 Mar 02 '20 at 12:48
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As you said, by writing out the values. – Dietrich Burde Mar 02 '20 at 12:48
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Chinese remainder theorem proves there is one answer modulo $40$ for each pair of values. – saulspatz Mar 02 '20 at 12:49
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I just used CRT. Is it okay to consider it as two equations and just use $\pm$ throughout? I got the right answer. – Loobear23 Mar 02 '20 at 12:50
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Perhaps you could include all your steps in the question. Otherwise people can't really say whether your method was OK, all they can do is show their own solutions. – David K Mar 02 '20 at 13:24
2 Answers
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1. case
If $p=5a+1$ and $p=8b-1$ then $5a+1=8b-1$ so $$5\mid 8b-2\implies 5\mid 16b-4\implies 5\mid b+1$$
so $b+1=5c$ and thus $$p= 8(5c-1)-1 = 40c-9$$ i.e. $$ p\equiv-9 \pmod{40}$$
Similary you do the other cases...
nonuser
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Using Chinese Remainder Theorem, you can prove that since $\gcd(8, 5) = 1$, for each value $a \mod 8$ and $b \mod 5$, you will always have unique $c\mod 40$.
1) $p = 1 \mod 8, p = 1 \mod 5$ yields $p = 1 \mod 40$
2) $p = 7 \mod 8, p = 1 \mod 5$ yields $p = 31 \mod 40$
3) $p = 1 \mod 8, p = 4 \mod 5$ yields $p = 9 \mod 40$
4) $p = 7 \mod 8, p = 4 \mod 5$ yields $p = 39 \mod 40$
Gratus
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