This is in continuation of
Get one half by squaring.
Consider this equation:
1/4 = (A1)-2 + (A2)-2 +.....+ (An)-2
where, each of the Ai's is a different positive integer - for all i=1 to n (inclusively).
n must be greater than 1.
- Determine the minimum value of n, for which there is at least one solution. Provide at least one valid representation for that smallest value.
- Determine the next three possible values of n for which there is a valid solution (At least one representation corresponding to each of the three values of n must be given.)
- Determine the maximum value of n that is achievable by way of simulation.
Only as a bonus question, display all possible representations corresponding to the minimum value of n.
a1 cannot be 2 as the problem forbids this. So a1>2.
Assume a1 is not 3 or 4. Then:
For n=2, the largest possible sum is 1/5^2+1/6^2, or 0.0671, a contradiction.
For n=3, the largest possible sum is 1/5^2+1/6^2+1/7^2, or 0.0881, a contradiction.
For n=4, the largest possible sum is 1/5^2+1/6^2+1/7^2+1/8^2, or 0.1038, a contradiction.
For n=5, the largest possible sum is 1/5^2+1/6^2+1/7^2+1/8^2+1/9^2, or 0.1161, a contradiction.
For n=6, the largest possible sum is 1/5^2+1/6^2+1/7^2+1/8^2+1/9^2+1/10^2, or 0.1261, a contradiction. Furthermore, the infinite sum of all larger squares is 0.0952, (sum 1 to infinity 1/(n+10)^2= 0.0952) so no number of additional squares will sum to 1/4.
Accordingly a1 is less than 5.
Say n=5, and a1=3. Then 1/3^2+1/5^2+1/6^2+1/7^2+1/8^2= 0.2149, insufficient. a1=4 must be still less, so for n=5, let both a1=3 and a2=4. Then we have the new problem (a3,..a5) = 11/144, or 0.0763. Assume a3 is at least 6. Then 1/6^2+1/7^2+1/8^2 = 0.0638, which is insufficient.
So, say a1=3, a2=4, a3=5. Then since 1/3^2+1/4^2+1/5^2+1/7^2+1/8^2 is 0.2496, insufficient, the sole remaining candidate is a1=3, a2=4, a3=5, a4=6, a5=7, but this does not equal 1/4.
So n is at least 6.
Say a1=4 and n=6, then we have the new problem (a2,a3,..a6) = 3/16, or 0.1875. But 1/5^2+1/6^2+1/7^2+1/8^2+1/9^2, or 0.1161, which is insufficient.
Say a1=3 and n=6, then we have the new problem (a2,a3,..a6) = 5/36, or 0.1388. Again 1/5^2+1/6^2+1/7^2+1/8^2+1/9^2 is insufficient.
So let both a1=3 and a2=4, then we have the new problem (a3,..a6) = 11/144, or 0.0763. Assume a3>5, then 1/6^2+1/7^2+1/8^2+1/9^2 = 0.0761, close but insufficient. So let a3=5. Then we have the new problem (a4,..a6)
=131/3600, or 0.0363. 1/9^2+1/10^2+1/11^2 = 0.0306, insufficient, so a4=6 or 7 or 8.
Say a4=6. Then we have the new problem (a5,a6) = 31/3600, but this has no integer solutions.
Say a4=7. Then we have the new problem (a5,a6) = 2819/176400 but this has no integer solutions.
Say a4=8. Then we have the new problem (a5,a6) = 299/14400 but this has no integer solutions.
Say a4=6, and a5=7, but that is too large.
Say a4=6, and a5=8. but that is too large.
Say a4=7, and a5=8. Then we have the new problem a6 = 251/705600, but this has no integer solution.
So n is at least 7. Clearly, this method, though exhaustive, becomes exponentially more tedious as n increases. Using the above results I was however able to derive (see note) a solution for n=9, which I conjecture is minimal n for this puzzle.
(1/3^2+1/4^2+1/5^2+1/6^2+1/12^2+1/30^2+1/60^2+1/75^2+1/100^2)
Since 1/12^2=1/15^2+1/20^2, clearly n=10 is a solution.
Since 1/60^2= 1/65^2+1/156^2, clearly n=11 is also a solution.
Since 1/156^2= 1/195^2+1/260^2, clearly n=12 is also a solution.
Note that, e.g. 1/(12(4z+1))^2=1/(60z+15)^2+1/(80z+20)^2, so it's not obvious to me why there should be a maximum value of n.
Note 2. My conjecture is false. n can be 8: 1/3^2+1/4^2+1/5^2+ 1/7^2+1/8^2+ 1/56^2 +1/168^2+1/840^2, a solution I just found under the original puzzle referenced above. I could have saved myself quite some time!
Edited on September 29, 2022, 3:29 am
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Posted by broll
on 2022-09-28 10:25:43 |
The story so far
