# THE NUMBER OF ORDERED PAIRS OF INTEGERS (x,y) SATISFYING THE EQUATION x^2+6x+y^2

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Rather than solving this question directly, i will let u know how these questions are solved taking some different example..

The pairs (x,y) of positive integers such that x^2-y^2 = 275 correspond 1-to-1 to pairs of positive factors (a,b) of 275 with a > b. Indeed, given (x,y) positive with x^2-y^2 = 275, we factor:

(x+y)(x-y) = 275

Clearly x > y, as x^2 = y^2+275 > y^2. So x+y > x-y > 0, and letting a = x+y, b = x-y, we have the desired pair of factors.

Conversely, factor 275 = ab, with a, b positive and a > b. Let x = (a+b)/2, y = (a-b)/2. These are positive integers since a and b must both be odd and hence their sum and difference must be even. Check that, indeed, x^2-y^2 = 275. Furthermore, check that these two maps are inverse to each other.

So now we just need to find the number of pairs of factors of 275. Well, let's just factor 275.

275 = 5^2 * 11

So the number of factors of 275 is the number of factors of 5^2 times the number of factors of 11, which is 3*2 = 6. We're looking for certain ordered pairs, so divide by 2, and we get 3 pairs. We can list these out: (275,1), (55,5), (25,11). Use the correspondence above to find the values for x and y, e.g. x = (275+1)/2, y = (275-1)/2.

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All the best.

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