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Old 2017-02-25, 23:20   #3
science_man_88
 
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Jul 2009
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Quote:
Originally Posted by MattcAnderson View Post
Hi Mersenneforum,

In order to tackle this first number theory problem,
(okay it is problem 84)
I will try a similar problem.
Find all x such that
5*x+6 is congruent to 0 mod 7 expression 1

From the reading, I notice that the greatest common divisor of 5 and 6 is 1. So the techniques presented here should apply.

We make an augmented T table

X 5*X 5*X mod 7
______________________
0 0 0
1 5 5
2 10 3
3 15 1
4 20 6
5 25 4
6 30 2

From expression 1, conclude that 5*x is congruent to 1 mod 7.
My education has a gap here.

I use my Maple computer tool to conclude that x is congruent to 2*x+1 mod 7.
Here is my Maple code –

a:=-(5*x+6) mod 7

And, again, Maple returns a= 2*x+1 mod 7.
Maybe someone can help me fill in the gaps.
Regards,
Matt
5x+6 \equiv 0 \pmod 7\\<br />
5x\equiv -6 \pmod 7 \\<br />
5\equiv -2 \pmod 7 \\ <br />
\therefore  \\<br />
5x\equiv -2x \pmod 7 \\<br />
-6 \equiv -2x\pmod 7 \\<br />
3 \equiv x \pmod 7  \\<br />

but this equation isn't quadratic.

edit:

the solution like this for 84 is similar to this:

\text {deleted previous ramble after checking with PARI/GP} \\<br />
12x+24 \equiv 0 \pmod {33} \\<br />
3(4x+8)\equiv 0 \pmod {3(11)}\\<br />
(4x+8)\equiv 0 \pmod {11}\\<br />
4(x+2)\equiv 0 \pmod {11}\\<br />
\text { by 0 product rule one of these has to be 0 and 4 is not 0} \\<br />
x+2 \equiv 0 \pmod {11}\\<br />
x\equiv -2 \pmod {11}\\<br />
x \equiv 9 \pmod {11}\\<br />

Last fiddled with by science_man_88 on 2017-02-25 at 23:41
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