Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation Links

 Math Help −> Puzzles −> Radical Thinking 

Radical Thinking

For given positive integers a and n, show there is a positive integer b such that

(sqrt(a) - sqrt(a-1))n = sqrt(b) - sqrt(b-1)

Source: http://www.math.purdue.edu/pow/spring2008/pdf/solution5.pdf 

Solution

Start by expressing 2 sqrt(b) in terms of sqrt(b) - sqrt(b-1), and then, assuming such an integer, b, exists for which

(sqrt(a) - sqrt(a-1))n = sqrt(b) - sqrt(b-1),

express b in terms of (sqrt(a) - sqrt(a-1))n.  Here we go:

2 sqrt(b) = (sqrt(b) + sqrt(b-1)) + (sqrt(b) - sqrt(b-1))

   = 1/(sqrt(b) - sqrt(b-1)) + (sqrt(b) - sqrt(b-1))

   = 1/(sqrt(a) - sqrt(a-1))n + (sqrt(a) - sqrt(a-1))n 

   = (sqrt(a) + sqrt(a-1))n + (sqrt(a) - sqrt(a-1))n 

4b = ( (sqrt(a) + sqrt(a-1))n + (sqrt(a) - sqrt(a-1))n )2 

From the binomial formula,

(sqrt(a) + sqrt(a-1))n = sum from k=0 to n of C(n,k) sqrt(a)n-k sqrt(a-1)k 

(sqrt(a) - sqrt(a-1))n = sum from k=0 to n of (-1)k C(n,k) sqrt(a)n-k sqrt(a-1)k 

Summing and then squaring the two equations, above, the left side is 4b, and on the right side notice that the odd k's cancel, so

if n is even, 4b = (sum from k=0 to n/2 of 2 C(n,2k) an/2-k (a-1)k)2 = 
          = 4 (sum from k=0 to n/2 of C(n,2k) an/2-k (a-1)k)2 

if n is odd, 4b = (sum from k=0 to (n-1)/2 of 2 C(n,2k) an/2-k (a-1)k)2 = 
          = 4a (sum from k=0 to (n-1)/2 of C(n,2k) a(n-1)/2-k (a-1)k)2 

In either case, both sides can be divided by 4, resulting in a formula for b which is an integer.

More about these sequences

Using Mathematica to solve this equation for general n proved fruitless; however, for specific n's the solutions were

If n=1, b=a=a(1)^2
If n=2, b=4a^2-4a+1=(1-2a)^2
If n=3, b=16a^3-24a^2+9a=a(3-4a)^2
If n=4, b=64a^4-128a^3+80a^2-16a+1 = (1-8a+8a^2)^2
If n=5, b=256a^5-640a^4+560a^3-200a^2+25a = a*(5-20a+16a^2)^2
If n=6, b=1024a^6-3072a^5+3456a^4-1792a^3+420a^2-36a+1 = (1-18a+48a^2-32a^3)^2
If n=7, b=4096a^7-14336a^6+19712a^5-13440a^4+4704a^3-784a^2+49a = a(7-56a+112a^2-64a^3)^2
If n=8, b=16384a^8-65536a^7+106496a^6-90112a^5+42240a^4-10752a^3+1344a^2-64a+1 = (1-32a+160a^2-256a^3+128a^4)^2
If n=9, b=65536a^9-294912a^8+552960a^7-559104a^6+329472a^5-114048a^4+22176a^3-2160a^2+81a = a(9-120a+432a^2-576a^3+256a^4)^2
If n=10, b=262144a^10-1310720a^9+2785280a^8-3276800a^7+2329600a^6-1025024a^5+274560a^4-42240a^3+3300a^2-100a+1 = (1-50a+400a^2-1120a^3+1280a^4-512a^5)^2

Now, making triangles out of these bold-face expressions,

. . . . . .  put this into a triangle: 

Even n's
0 1
2 1 -2
4 1 -8 8
6 1 -18 48 -32
8 1 -32 160 -256 128
10 1 -50 400 -1120 1280 -512

 

Odd n's
1 1
3 3 -4
5 5 -20 16
7 7 -56 112 -64
9 9 -120 432 -576 256

Dividing each column by successive powers of 4, and then searching in OEIS for triangles of these coefficients, I came upon Riordan Arrays A110162 and A111125 . . . . . . (make good OEIS references, and maybe something about Riordan Arrays -- what are they?)

So then the Even arrays =(mycombin($A37+B$36,$A37-B$36)+mycombin($A37+B$36-1,$A37-B$36-1))*2^(2*B$36-1)*(-1)^B$36

and the odd arrays =mycombin($A46+B$45,$A46-B$45)*(2*$A46+1)/(2*B$45+1)*2^(2*B$45)*(-1)^B$45

. . . . . .  fix these to give good values for n.

Refer to PuzzlePurdue-2008a-05-SqrtIdentity.xls

Internet references

xxx

Related pages in this website

xxx

 

The webmaster and author of this Math Help site is Graeme McRae.