Hello, Petrus!

Here's a rather primitive solution.

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- September 17th 2012, 05:09 AM #1

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- September 17th 2012, 05:45 AM #2

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- September 17th 2012, 05:55 AM #3

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- September 17th 2012, 07:47 AM #4

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## Re: need help!

Well, there is always "brute strength"- 4(1)= 4, 4(2)= 8, 4(3)= 12, 4(4)= 16= 2 (mod 14), 4(5)= 20= 6 (mod 14)- Aha! But we don't want to stop there:

4(6)= 24= 10 (mod 14), 4(7)= 28= 0 (mod 14), 4(8)= 32= 4 (mod 14), 4(9)= 36= 8 (mod 14), 4(10)= 40= 12 (mod 14), 4(11)= 44= 2 (mod 14), 4(12)= 48= 6 (mod 14)!!, 4(13)= 52= 10 (mod 14). So we have**two**solutions, mod 14: 5 and 12. Of course, because those are modulo 14, the equation is satisfied by those plus any multiple of 14: 5+ 14i, and 12+ 14j.

The first, 5+ 14i is is equivalent to Soroban's 7k- 2 for k odd and 12+ 14j to 7k- 2 for k even.

A bit more formal method is to say the 4x= 6 (mod 14) means that there exist an integer n such that 4x= 6+ 14n which is the same as 2x= 3+ 7m or 2x- 7n= 3. Now, because 2 and 7 are "relatively prime" we can use Euclid's algorithm: 2 divides into 7 three times with remainder 1 so (1)7- (3)2= 1. Multiplying both sides by 3, (3)7- (9)2= 2(-9)- 7(-3)= 3. That is, x= -9 and n= -3 is a solution to that equation. Of course, x= -9+ 7i, n= -3+ 2i also satisfies that equation: 2(-9+ 7i)- 7(-3+ 2i)= -18+ 14i+ 21- 14i= 3 for all i. Replacing "i" with "k+ 1" gives x= -9+ 7(k+ 1)= -9+ 7k+7= -2+ 7k, Soroban's answer.