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If a and b are non-zero integers, then there exist four integers h,k,r,s such that hs-kr =1,ak+bs=0

Parashara K.V.R , 13 Years ago

Grade 9

2 Answers

Arun Kumar

Hi
$\\hs-kr =1 \\ak+bs=0 \\$we have to find 4 such integers$ \\ak=-bs \\=>k={-bs \over a} \\=>$if $b=a*n_1$ then no problem$ \\$so for the problem we will assume $b \neq a*n \\=>s=n_1*a \\=>k=n_1*b \\$we got s,k$$
$\\$for any equation of form ax+by=c$ \\$we can integral solution only when GCD(a,b) divides c$ \\so \\hs-kr=1 \\d=(s,r)=n_1\, divides\, 1=>n_1=1 \\$let's say h_0, k_0 are 1 pair of solution others pair we can find by $ \\h=h_0+{rn \over gcd(s,r)},k=k_0-{kn \over gcd(s,r)}$
but h0 and k0 we have to find by hit and trial by lemma its sure that it will be found given condition is satisfied.
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Last Activity: 11 Years ago

mycroft holmes

Consider the fraction a/b. If it is not already in reduced form then consider the equivalent fraction x/y such that gcd(|x|, y) =1

Then bx + a(-y) = 0. Also, since gcd (|x|, y) =1, there exist integers m, n such m|x|+ny =1

If x>0, then mx – n(-y) =1 so that we can choose h=m, k=-y, r=n, s=-y

If x<0, then (-m)x -n (-y) =1 so that we can choose h = -m, k=-y, r=n, s=-y

Thus we can always find h,k,r,s satisfying the given conditions

Last Activity: 11 Years ago

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