Problem 1

Find all integers a,b,c satisfying

1<a<b<c

such that

(a-1)(b-1)(c-1)

is a divisor of

abc-1

Step 6 of 6: Exclude a=3 and a=4 for q≥3

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a=3\text{ or }4,\ q\ge3\ \Longrightarrow\ (2q-3)bc>2q(b+c)\text{ or }(3q-4)bc>3q(b+c)

Detailed analysis

For a=3 the rearranged equation is (2q−3)bc+(2q+1)=2q(b+c). Since b≥4, the left product term is at least (8q−12)c, which is greater than 2q(b+c). For a=4, the equation is (3q−4)bc+(3q+1)=3q(b+c); b≥5 gives an even stronger strict inequality. Both cases are impossible. Therefore the only solutions are (2,4,8) and (3,5,15), and direct substitution verifies both.