Try this beautiful problem from the PRMO, 2018 based on Smallest value.
Let a and b natural numbers such that 2a-b, a-2b and a+b are all distinct squares. What is the smallest possible value of b?
Algebra
Numbers
Multiples
Answer: is 21.
PRMO, 2018, Question 15
Higher Algebra by Hall and Knight
2a-b=\(k_1^2\) is equation 1
a-2b=\(k_2^2\) is equation 2
a+b=\(k_3^2\) is equation 3
adding 2 and 3 we get
2a-b=\(k_2^2+k_3^2\)
or, \(k_2^2+k_3^2\)=\(k_1^2\) \((k_2<k_3)\)
For least 'b' difference of \(k_3^2\) and \(k_2^2\) is also least and must be multiple of 3
or, \(k_2^2\)=a-2b=\(9^2\) and \(k_3^2\)=a+b=\(12^2\)
or, \(k_3^2-k_2^2\)=3b=144-81=63
or, b=21
or, least b is 21.
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