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

```
9 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.

All the best.
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Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
9 years ago
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