2015 can be written as the difference of the squares of two positive

Question

2015 can be written as the difference of the squares of two positive integers in how many ways?

A. 1
B. 2
C. 3
D. 4
E. 5

globaldesi · Answer

can you please follow the guidelines while posting a question.
Ideally your question should have options and timers
Otherwise the solution would be something 
 a^2 - b^2  = 2015
or
(a+b)(a-b) = 13*5*31
the solution would be to form pairs 
a+b= 65; a-b = 31
a+b= 155; a-b= 13
a+b = 403; a-b = 5
thus 3 such pairs exists
for each value of a and b
Please let me know if my answer is correct

rvgmat12 · Answer

what about a + b = 2015 and a - b = 1

globaldesi · Answer

Yes, I think that also . Apologies missing a lot of edge cases these days

ScottTargetTestPrep · Answer

Solution:
 
We want to express 2015 as x^2 - y^2 where x and y are positive integers with x > y. In other words, we want to solve the equation x^2 - y^2 = 2015 where x and y are positive integers. Notice that x^2 - y^2 can be factored as (x - y)(x + y). Therefore, we need to express 2015 as a product of two positive integers:

2015 = 1 * 2015 = 5 * 403 = 13 * 155 = 31 * 65

If we take the first product, we can create the equation:

(x - y)(x + y) = 1 * 2015

Since x - y > x + y, then x - y = 1 and x + y = 2015. If we add x - y = 1 and x + y = 2015, we have:

2x = 2016

x = 1008

Since x - y = 1, then y = 1007. Therefore, a pair of positive integers that satisfy the equation x^2 - y^2 = 2015 is (x, y) = (1008, 1007). Recall that 2015 can be written as a product of two positive integers in 3 more ways. So, if we create the following equations:

(x - y)(x + y) = 5 * 403, (x - y)(x + y) = 13 * 155, and (x - y)(x + y) = 31 * 65

Solving these equations as we did for (x - y)(x + y) = 1 * 2015, we will find 3 more distinct pairs of positive integers that satisfy the equation x^2 - y^2 = 2015. Therefore, there are 4 solutions to the equation x^2 - y^2 = 2015 with x and y being positive integers.

Answer: D

Rohit5678 · Answer

One way to avoid this silly mistake is to count the no. of factors:

(a+b)(a-b) =2015 =  13^1*5^1*31^1

So no. of factors = (1+1)(1+1)(1+1) = 2*2*2 = 8

So we have 4 pairs of nos that get the product

av1901 · Answer

Another approach to this problem can be that once we find the prime factorization of 2015 = 5*13*31
We can say that total number of factors = 2*2*2 = 8
So, there will be 4 factor pairs which can be expressed as a^2-b^2 or (a+b)(a-b)

D

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2015 can be written as the difference of the squares of two positive

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