Show that for all positive integers n hcf6n 8 4n 5 1 Supp

Show that for all positive integers n, hcf(6n + 8, 4n + 5) = 1. Suppose a, b are integers such that a | b and b | a. Prove that a = plusminus b. Suppose s, t, a, b are integers such that sa + tb = 1. Show that hcf(a, b) = 1.

Solution

(a)  Let the hcf = k
k is a positive integer
Then k divides 6n + 8 and 4n + 5
2(6n + 8) = 12n + 16
3(4n + 5) = 12n + 15
So k divides 12n + 16 and 12n + 15
(12n + 16) - (12n + 15) = 1
So k divides 1.
k = 1.

(b)  If ab then b=na, and ba then a=mb,

thus: ab = mnab

=> mn= 1 m = n = 1 or m = n = 1

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