My first problem
Number Theory
Level
4
Find the sum of all possible positive integers that
is a perfect square.
The answer is 4.
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1 solution
If the above equation is to equal a perfect square, then let us write:
2^(2n-1) - 2^n + 1 = a^2;
or 2*2^(2n-2) - 2^n + 1 = a^2;
or 2 2^(2n-2) - 2 2^(n-1) + 1 = a^2;
or 2^2(n-1) + 2^2(n-1) - 2*2^(n-1) + 1 = a^2;
or [2^(n-1)]^2 - 2*2^(n-1) + 1 = a^2 - [2^(n-1)]^2;
or [2^(n-1) - 1]^2 = [a + 2^(n-1)][a - 2^(n-1)].
Now, let us now equate the terms:
a - 2^(n-1) = 1 (i); a + 2^(n-1) = [2^(n-1) - 1]^2 (ii).
whereby solving (i) for 2^(n-1) in terms of a yields: a-1 = 2^(n-1), and substituting this value into (ii) produces:
a + (a-1) = [(a-1) - 1]^2 => 2a - 1 = (a - 2)^2 => 2a - 1 = a^2 - 4a + 4 => 0 = a^2 - 6a + 5 => 0 = (a-1)(a-5) => a = 1 or 5.
A quick check in (i) produces a = 1 => 2^(n-1) = 0 and a = 5 => 2^(n-1) = 4, which only the second choice is valid and ultimately yields n = 3. Next, let us examine the form:
2^(2n-1) - 2^n = a^2 - 1;
or (2^n)*[2^(n-1) - 1] = (a+1)(a-1)
and equate the larger and the smaller term respectively as:
2^n = a + 1 (iii) 2^(n-1) - 1 = a - 1 (iv)
or 2^n = 2a after simplifying (iv). Substituting this expression into (iii) ultimately produces: 2a = a + 1 => a = 1 and finally n = 1.
Thus, the only values of n that can ever produce a perfect square of the form 2^(2n-1) - 2^n + 1 are n = 1, 3 which sum to 4.