Problem 1:

Given two integers a and m larger than 1, show that, if m is odd, then then is a divisor of . Use this result to obtain the factorization of 1001.

Solution 1:

Since m is odd, we have and the result follows.

This shows that

.

Problem 2:

Generalize the result of the above problem to obtain that if a and m are two integers larger than 1 and if is an odd divisor of m, then is a divisor of . Use this result to show that 101 is a factor of 1000001.

 Solution 2:

The result is immediate if we write and then we apply the result of the above problem. This shows that

.

Problem 3:

Show that 7, 11 and 13 are factors of

Solution 3:

It follows from the previous result that .

The result then follows from the fact that 7, 11, and 13 factors of 1001.

Problem 4:

Show that is a composite number for each integer . More generally, show that if a is a positive integer such that is a perfect square, then is a composite number provided that

 Solution 4:

It is enough to observe that . For the general case, we only need to observe that .

We also make an observation that the condition is sufficient but not necessary.

Cheers,

Nalin Pithwa