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

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

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

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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

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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