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 1 of 5: Reduce the case b<a

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b<a,\ (a-1)\mid(b-1),\ 0\le b-1<a-1\ \Longrightarrow\ b=1

Detailed analysis

If $a=1$ , the definition of multiple forces $b-1=0$ , so $(a,b)=(1,1)$ . Now assume $a>1$ . If $b<a$ then $0\le b-1<a-1$ , and since $a-1$ divides $b-1$ the only multiple of $a-1$ in that range is $0$ , so $b=1$ . Every pair $(a,1)$ indeed satisfies both conditions, since $b^2=1$ trivially divides $a^3$ and $b-1=0$ is a multiple of $a-1$ .