Find all solutions to the equation 2a2b2 where a b are posit

Find all solutions to the equation 2^a-2^b=2 where a, b are positive integers

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Ans-

. Let n 1 be an integer. What is the maximum number of disjoint pairs of elements of the set {1, 2, . . . , n} such that the sums of the different pairs are different integers not exceeding n? Solution. Consider x such pairs in {1, 2, . . . , n}. The sum S of the 2x numbers in them is at least 1+2+· · ·+2x since the pairs are disjoint. On the other hand S n+(n1)+· · ·+(nx+1) because the sums of the pairs are different and do not exceed n. This gives the inequality 2x(2x + 1) 2 nx x(x 1) 2 , which leads to x 2n1 5 . Hence there are at most 2n1 5 pairs with the given properties. We show a construction with exactly 2n1 5 pairs. First consider the case n = 5k + 3 with k 0, where 2n1 5 = 2k + 1. The pairs are displayed in the following table. Pairs 3k + 1 3k · · · 2k + 2 4k + 2 4k + 1 · · · 3k + 3 3k + 2 2 4 · · · 2k 1 3 · · · 2k 1 2k + 1 Sums 3k + 3 3k + 4 · · · 4k + 2 4k + 3 4k + 4 · · · 5k + 2 5k + 3 The 2k+1 pairs involve all numbers from 1 to 4k+2; their sums are all numbers from 3k+3 to 5k + 3. The same construction works for n = 5k + 4 and n = 5k + 5 with k 0. In these cases the required number 2n1 5 of pairs equals 2k + 1 again, and the numbers in the table do not exceed 5k + 3. In the case n = 5k + 2 with k 0 one needs only 2k pairs. They can be obtained by ignoring the last column of the table (thus removing 5k + 3). Finally, 2k pairs are also needed for the case n = 5k + 1 with k 0. Now it suffices to ignore the last column of the table and then subtract 1 from each number in the first row. Comment. The construction above is not unique. For instance, the following table shows another set of 2k + 1 pairs for the cases n = 5k + 3, n = 5k + 4, and n = 5k + 5. Pairs 1 2 · · · k k + 1 k + 2 · · · 2k + 1 4k + 1 4k 1 · · · 2k + 3 4k + 2 4k · · · 2k + 2 Sums 4k + 2 4k + 1 · · · 3k + 3 5k + 3 5k + 2 · · · 4k + 3 The table for the case n = 5k + 2 would be the same, with the pair (k + 1, 4k + 2) removed. For the case n = 5k + 1 remove the last column and subtract 2 from each number in the second row