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