(a) prove that for all n belongs to N, gdc(2n+1, 9n+4)=1

(b) prove that for all n belongs to N, gdc(5n+8, 3n+5)=1

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(a) gdc(2n+1, 9n+4) =

= gdc(2n+1, 9n+4 - 4*(2n+1))

=gdc(2n+1,n)

= gcd(2n+1,2n)

= 1

(b) gdc(5n+8, 3n+5)=

= gdc(5n+8, 2*(3n+5) -(5n+8))

= gdc(5n+8, n+2)

= gdc(5n+8, 5(n+2))

= gdc(5n+8, 5(n+2) - (5n+8))

= gdc(5n+8, 2)

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