Find all integers n for which
√(25/2 + √(625/4 –n)) + √(25/2 - √(625/4 –n)) is an integer.
assumed
n value computed integer value
0 5.0 5
144 7.0 7
784 8.9999999999999999998 9
2304 10.9999999999999999998 11
5184 12.9999999999999999998 13
10000 14.9999999999999999998 15
17424 16.9999999999999999998 17
28224 18.9999999999999999998 19
43264 20.9999999999999999998 21
63504 22.9999999999999999998 23
90000 24.9999999999999999998 25
from
4 kill "rtsumeqi.txt"
5 open "rtsumeqi.txt" for output as #2
10 for N=-10000 to 100000
20 V=sqrt(25//2+sqrt(625//4-N))+sqrt(25//2-sqrt(625//4-N))
30 if im(V)<0.000001 then
40 :Vr=int(re(V)+0.5)
50 :if abs(re(V)-Vr)<0.000001 then
60 :print N,V,Vr
65 :print #2,N,V,Vr
70 next
80 close
The values of the formula are just the odd numbers starting with 5. The values of n start with zero and appear to be the square of a quadratic polynomial as the square roots appear to increase by a linearly increasing amount.
144 = 12^2
784 = 28^2 16
2304 = 48^2 20
5184 = 72^2 24
10000 = 100^2 28
17424 = 132^2 32
28224 = 168^2 36
43264 = 208^2 40
63504 = 252^2
90000 = 300^2
If p is the position in the sequence, starting with position zero:
sqrt(n) = ap^2 + bp + c
0 = c
12 = a + b
28 = 4a + 2b
2a = 4
a=2
b=10
n = (2*p^2 + 10p)^2 for p = 0 on up
v = 2*p + 5
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Posted by Charlie
on 2015-03-10 14:21:27 |
with the assistance of a computer language that can handle complex intermediate terms
