Question

Let a and b be natural numbers such that 2a-b, a-2b and a+b are all distinct squares.
What is the smallest possible value of b?

Answer 1

Step-by-step explanation:

a-2b is the smallest possible value of b

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Answer 2

If you subtract the last two you get q2−p2=3b. If you add the first and last you get k2+q2=3a. No primitive Pythagorean triangle has legs that differ by a multiple of 3, so we need a triangle that has a common factor of 3. The smallest such is 9−12−15 and we find

3b=144−81=63b=213a=225+144=369a=123

This is the smallest b because the difference of the two legs must be at least 3. If the shorter leg is c we have (c+3)2−c2=6c+9 and b will grow with the shorter leg