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Question

The number of different positive integer triplets (x,y,z) satisfying the equations x2 + y - z = 100 and  x + y2 - z = 124 is



A

0

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B

1

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C

2

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D

3

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Solution

The correct option is B

1


From the two given equations, on subtraction, we get

y2 - x2 + x - y = 24,

( y - x ) ( y + x - 1 ) = 24.

As x,y,z are positive integers, x + y - 1 > 1

y - x > 0 or y > x.

Since y + x and y - x are of the same parity (i.e., either both odd or both even ) y + x - 1  and y - x are of opposite parity. Also ( y + x - 1) - ( y - x ) = 2x - 1 > 0. Hence y + x - 1 > y - x .

Consider (i) gives x = 3  and y = 6. These values of x and y substituted in the given equations give two different negative values of z. hence case (i) is not admissible.

Case (ii) give x = 12, y = 13. Then from both the given equations we get z = 57. Thus there is a unique


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