Lots of clues
The following informations are supplied -
(i) We have a positive integer m .
(ii) The number 2 m + 1 doesn't have any factor which is a perfect square.
(iii) The sum of 2 m + 1 consecutive positive integers is the square of a positive integer p .
(iv) the starting member, a , of those integers, is the minimum possible.
Using these, find the value of 2 a − p .
The answer is 1.
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1 solution
Curious whether there are any solutions with higher values of m . I've hardly shown this solution is unique.
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Let the first of the 2 m + 1 consecutive positive integers be a . Then the sum is ( 2 m + 1 ) ( a + m ) . Since this is a perfect square and 2 m + 1 doesn't have any square factor, therefore a = m + 1 , 7 m + 4 , 1 7 m + 9 , . . . . The minimum value of a is m + 1 , p = 2 m + 1 and hence 2 a − p = 1 . As an example, 8 + 9 + 1 0 + 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + 1 9 + 2 0 + 2 1 + 2 2 = 2 2 5 = 1 5 2 . Here a = 8 , m = 7 , p = 1 5 , 2 a − p = 1 6 − 1 5 = 1 .
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Nice.
Oh, apparently Brilliant doesn't think "Nice" is sufficient.
Very nice.
"We have a positive integer m ..."
Ok, let's see if m = 1 works...
"The number 2 m + 1 doesn't have any factor which is a perfect square."
Well, 2 m + 1 = 3 , which is prime.
"The sum of 2 m + 1 consecutive positive integers is the square of a positive integer p ."
What does this mean? Does this refer to the sum of any 2 m + 1 consecutive positive integers? That cannot be. Let's keep reading.
"The starting member, a of those integers is the minimum possible."
OK, that clarifies the previous point. Let's start with a = 1 . Hmm... 1 + 2 + 3 = 6 . That doesn't work. Let's try a = 2 . We see 2 + 3 + 4 = 9 = 3 2 . So to minimize a and p we can take a = 2 and p = 3 . In which case 2 a − p = 1 .
Is a = 2 necessarily minimal? Yes, because if we try a = 1 , then the sum would be ∑ i = 1 k i = 2 k ( k + 1 ) , for k = 2 m + 1 . This is clearly not a perfect square. No arithmetic sum starting with a = 1 with more than one term will be a perfect square.