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 Root sum equals integer (Posted on 2015-03-10)
Find all integers n for which √(25/2 + √(625/4 –n)) + √(25/2 - √(625/4 –n)) is an integer.

 No Solution Yet Submitted by K Sengupta No Rating

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 with the assistance of a computer language that can handle complex intermediate terms | Comment 2 of 7 |
`                                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

 Posted by Charlie on 2015-03-10 14:21:27

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