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Progression of a Square (Posted on 2025-02-23) Difficulty: 3 of 5
For which of the following expressions is there no value of n>1 for which the given sum is a square?:

1+2+3+...+n
1+3+5+...+(2*n-1)
1+4+7+...+(3*n-2)
1+5+9+...+(4*n-3)
1+6+11+...+(5*n-4)
1+7+13+...+(6*n-5)
1+8+15+...+(7*n-6)
1+9+17+...+(8*n-7)

Also:

For which two of them are there values of n such that the sum is the perfect number 2,305,843,008,139,952,128?


From BENT Brain Ticklers, Winter 2024

See The Solution Submitted by Charlie    
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Possible Solution | Comment 1 of 2
For which of the following expressions is there no value of n>1 for which the given sum is a square?

Note: the 'sum' referred to is the sum of the first n terms in each sequence, not as I initially thought the sums, n, ((2n-1), etc.

The answer to the first part is that solutions exists for all except the last (8n-7). The sum of that series is 4n^2-3n, so its implied root is sqrt(n)sqrt(4n-3). To clear the first square root, n must be a square; but then to clear the second requires some number 3 less than a square also to be a square, but this can happen only if n=1.


Edited on February 23, 2025, 10:59 am
  Posted by broll on 2025-02-23 09:49:40

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