Problem 1

Find all pairs

(a,b)

of positive integers such that

a^3

is a multiple of

b^2

, and

b-1

is a multiple of

a-1

. (Here

n

is called a multiple of

m

n=km

for some integer

k

Step 3 of 5: Rule out c<a

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c<a\Rightarrow c=1\Rightarrow a=d^2,\ b=d^3,\quad (d^2-1)\mid(d^3-1)\Longrightarrow(d+1)\mid1\Longrightarrow d=1

Detailed analysis

If $c<a$ , the congruence $c\equiv1\pmod{a-1}$ with $0<c<a$ forces $c=1$ , so $a^3=b^2$ . Writing $a=d^2$ , $b=d^3$ for a positive integer $d$ , the divisibility $(a-1)\mid(b-1)$ becomes $(d^2-1)\mid(d^3-1)$ , i.e. $(d-1)(d+1)\mid(d-1)(d^2+d+1)$ . For $d>1$ this reduces to $(d+1)\mid(d^2+d+1)=d(d+1)+1$ , forcing $(d+1)\mid1$ , impossible. So $d=1$ , giving the trivial pair $a=b=1$ , already counted.