Impossible Solution


Given that a and b are positive integers, find the conditions for which the equation radicala minus b = radicalc has a solution.


From radicala = b + radicalc, square both sides, a = b2 + 2bradicalc + c.

Rearranging we get,
a minus b2 minus c
= radicalc.

As the left hand side is rational, radicalc must be rational.

Let radicalc=x/y, where HCF(x, y)=1.

Squaring, c=x2/y2, cy2=x2.

As the left hand side divides by y2 and HCF(x2, y2)=1, the right hand side will only divide by y2 if y2=1. Hence c=x2 must be a perfect square.

Furthermore, if c is a perfect square, radicala = b + radicalc will be integer, so a must also be a perfect square.

Problem ID: 190 (28 Nov 2004)     Difficulty: 3 Star

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