If two numbers are coprime, then they are natural and can be put in an equation so that they equal 1. But how?

You are supposed to be able to use the opposite to the euclidean algorithm, but i dont know how.If two numbers are coprime, then they are natural and can be put in an equation so that they equal 1. But how?
I'm not sure I follow the question.





However, to the other poster:


6*6 - 1*35 = 1.





For this example, you do the following:


35 = 5*6 + 5


6 = 1*5 + 1





which we may rewrite as:


35 - 5*6 = 5


6 - 1*5 = 1





Substituting the first equation into the second (for the 5), we get


6 - 1*(35 - 5*6) = 1


6 - 1*35 + 5*6 = 1





6*6 - 1*35 = 1If two numbers are coprime, then they are natural and can be put in an equation so that they equal 1. But how?
( I assume to begin with that you know how to work the Euclidean algorithm )





Take the example 147 梅 23 .... (the 梅 is a division sign; it doesn't show up well here.)








147 梅 23 = 6(23) + 9 ...... (A)





23 梅 9 = 2(9) + 5 .............(B)





.. 9 梅 5 = 1(5) + 4 ...........(C)





.. 5 梅 4 = 1(4) + 1 ...........(D)





... 4 梅 1 = 4(1) + 0





Now you work from the bottom up (the working is on the right; on the left is some explanation, which is normally omitted in doing the work)





From line (D) ............ ................ .......... 1 = 5 - 1(4)





From (C), 4 = 9 - 1(5) ......... ........... .........= 5 - (9 - 5) = 2(5) - 9





From (B), 5 = 23 - 2(9) ............. ........... ....= 2 { 23 - 2(9) } - 9 = 2(23) - 5(9)





From (A), 9 = 147 - 6(23) ............. .............= 2(23) - 5 { 147 - 6(23) }





............. ................. ............... .................= 32(23) - 5(147)





So 1 = 32(23) - 5(147)





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(But it works the same whatever the result of the Euclidean algorithm is. Starting with the greatest common factor and working back up the stack like that gives you an expression for the greatest factor in terms of the two numbers you started out with. If they are co-prime, then the greatest factor has to be 1, of course.)
The theorem is:


If a and b are co-prime, then integers p, q exist such that ap+bq = 1, and hence, if a and b are positive, positive integers P, Q exist such that mod(aP-bQ) =1.





I don't think it is difficult to prove, but you better consult a book, or maybe an internet source, for the proof.
35 and 6 are coprime ... I don't see how you'd put them in an equation to equal 1