Well this is yet another instance of a sequence where trigonometric substitutions can save us a lot of sweat. For those who came in late, the reason I say "yet another" are http://www.goiit.com/posts/list/trignometry-if-a-b-are-positive-quantities-and-if-41432.htm and http://www.goiit.com/posts/list/algebra-salutes-assured-41758.htm#205608.

 

and  

 

If we dispose of the trivial case  = = 0, This form should remind us of the formulas of cos(A+B) and sin(A+B).

 So, how to get these expressions into a trigonometric form. If we let  = r cos and  = r sin where r = ²+² then we can see that a₂ = r²(cos² - sin²) = r²cos(2). Similarly b₂ = r²(sincos+sincos) = r²sin(2).

In this way, we get a_n = rⁿcosn and b_n = rⁿsinn.

 

Hence a₁₉₉₇ = b₁ and b₁₉₉₇ = a₁ is the same as

 r¹⁹⁹⁷cos1997  = rsin and

r¹⁹⁹⁷sin1997  = rcos

 

Squaring and adding gives r = ²+²  = 1

 

Hence the eqns are cos1997  = sin 

                             sin1997  =  cos

This gives 1997 = 2n+/2 - or 1998 = 2n+/2

or   = 2n/1998+/3996

Since we are looking for distinct values of = cos and  = sin, we need only take values of n =1,2,3,...,1998.

 

So, the total number of distinct ordered pairs (,) including (0,0) is 1999.

 

While I was gloating over my solution I looked at the solution provided by the author which I have copied in the post below. I'll look quite a fool then!

                         

 

Even i am getting 360

 

Basically, the numbers r², s⁴ and t² have to appear in either number. This can be assigned in 2*2*2 = 8 ways.

 

Now if r² appears in one number, the other can have r⁰ or r¹ or r² as its factors.

 

This reasoning gives 8*3*5*3 = 360 such ordered pairs

Its a bit messy but alright:

 

Basically, the problem is to find the remainder when the number is divided by 100

 

(14)^{14^14} = (196)^{7^14} = (100n-4)^{7^14}. = 100n - 4^{7^14} (not the same n)

 

So, now we have to find the remainder when 4^{7^14} is divided by 100.

 

4⁷ = 16384 = 16400-16 = 100n-16

Hence the remainder when 4^{7^14} is divided by 100 is the same as when 16¹⁴ = 4²⁸ is divided by 100. This may look intimidating but it can be resolved quite simply.

4⁶ = 4096 = 51*100-4 = 100n-4.

4²⁸ = (4⁶)⁴ * 64

     = (100n-4)⁴ * 64

     = (100n+4⁴)*64

    = 100n+4⁸

    = 100n + (100k-4)*16

    = 100k-64

 

Summarizing, 14^{14^14} = 100n-4^{7^14}

                                = 100n-(100k-64)

                               = 100m+64

Hence the last two digits are 64

 

I said it would be messy

hey thanx for the compliment. (blush)