Question

True or False: If a|b and a|(b + 1), then a = ±1.

[NOTE: Use this Theorem 3: If a|b and a|c, then a|(bx+cy) for any integers x and y.
PROOF: Since a|b, there is an integer u such that b = au.
Since a|c, there is an integer v such that c = av.
Substituting, bx + cy = aux + avy = a(ux + vy)
since ux+ vy is an integer, a|(bx+cy). QED.]

I get stuck with my prove. I'm trying to following theorem 3.
MY ANSWER: True
Proof: Since a|b, there is integer u such that b=au
by theorem 3 tell us that a| b+1 - b. so we have a = ±1.
please help.

Answer 1

if a|b and a|c, then a|(b-c) (why?)
so,
if a|(b+1) and a|b,
a|(b+1-b)
a|1

Answer 2

thank you so much steve
i get it.
a|b, b=ae
a|b+1, b+1=af
substituting, ae+1=af
1=a(e-f), we conclude that a|b+1-b
therefore a|1, a = ±1