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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» Let 2a – b = n2 …(i) a – 2b = p2 …(ii) a + b = k2 …(iii) Adding (ii) and (iii) we get 2a – b = p2 – k2 p2 + k2 = n2 (p < k as a + b < a – 2b) For b to be smallest, k2 and p2 is also small and must be multiple of 3 (as 3b = k2 – p2) For smallest b, the least value of k and p be 12 and 9 resp. ∴ Least value of b = 21 |
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