11Base Step. Ifn=1, then4n+1+52n−1=16+5=21,which is obviously divisible by 21. Since our inductive step will involvetwo predecessors, we are obliged to check the casen=2. But 43+53=64+125=189=21×9.Inductive Step.4n+2+52n+1=4·4n+1+52·52n−1=4·4n+1+(4·52n−1−4·52n−1)+52·52n−1=4(4n+1+52n−1)+52n−1(52−4).Now the last term is divisible by 21; thefirst term, by the inductive hypoth-esis, and the second because 52−4=21.1.21For any integern≥2, prove that there arenconsecutive composite num-bers. Conclude that the gap between consecutive primes can be arbitrarilylarge.Solution.The proof has nothing to do with induction. If 2≤a≤n+1,thenais a divisor of(n+1)!; say,(n+1)! =dafor some integerd. Itfollows that(n+1)! +a=(d+1)a, and so(n+1)! +ais composite forallabetween 2 andn+1.1.22Prove that thefirst and second forms of mathematical induction are equiv-alent; that is, prove that Theorem 1.4 is true if and only if Theorem 1.12 istrue.Solution.Absent.1.23(
This is the end of the preview.
access the rest of the document.