DIRECTIONS for questions 17 to 25: Select the correct alternative from the given choices.
How many natural numbers less than 1000 can be expressed as the difference of two perfect squares in at least one way?
Answers
Step-by-step explanation:
The answer depends on whether you count 02 as a perfect square or not. For the analysis below, 02 is a perfect square. And we interpret "number below 100" as meaning positive integer below 100. Also, note that 5=32−22=(−3)2−22. Representations will always be non-unique if we allow sign changes. So we rule out using squares of negative integers.
The easy way to solve this problem is to look at 1 to 99, one after the other. For each, check whether it satisfies the condition, and keep a running tally. After a little while, we notice some patterns, and then the work is quick. As a partial compromise, we could look up a general result, and use that. But that is not interesting, so we develop all the theory we need.
Let n be a positive integer. We look for non-negative integer solutions of
x2−y2=n,that is, of(x−y)(x+y)=n.
We must then have
x−y=dx+y=nd
where d is a (positive) divisor of n.
Solving for x and y, we obtain
x=d+nd2andy=−d+nd2.
From the above equations, x and y are integers precisely when d and n/d are both even or both odd. And since y must be non-negative, we need to have d≤n/d, that is, d≤n−−√. Any factor d of n satisfying those conditions gives us a solution, and different factors yield different solutions.
Thus the number of ways of expressing n as a difference of 2 squares is precisely the same as the number of divisors d of n such that d≤n−−√ and d and n/d have the same parity. In particular, if n is divisible by 2 but not by 4, there is no representation of n.
Look first at the case n odd. There is only one factorization of n of the desired type iff n is an odd prime or n=1.
Look next at the case n even.
The number of representations of n is the number of ways of splitting n into two even factors d and n/d, with d≤n/d. This is the same as the number of ways of expressing n′=n/4 as a product of d′ and n′/d′, where d′≤n′/d′.
There is only one such factorization precisely if n′ is a prime or n′=1.
Now counting is easy, though somewhat tedious. For the odds, count 1, plus the odd primes ≤99. There are 24 odd primes less than 100, for a total of 25.
For the evens, count 1, plus all primes ≤24. There are 9 such primes, giving a total of 10 even numbers.
So overall 35 numbers qualify.
Comment: If we don't allow 02 as a perfect square, some changes need to be made. The most obvious ones are that 1 and 4 should not be counted. But the situation is more complicated than that. For the odd case, if p is a prime, then p2 now only has one representation, since p2−02 is not allowed, so it should be counted. Similarly, in the even case, now 4p2 only has one representation.