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  ### Math Virus Formula
  
  ```
  Date: 10/23/2001 at 05:49:48
  From: Victor Lee
  Subject: I cannot figure out the formula to a math virus
  
  Hello Dr. Math,
  
  I have to figure out a formula for a math virus. The virus spreads to
  all the squares directly touching each other (not including
  diagonally), and I have found the formula for the number of newly
  infected cells (although this does not include the first minute.) The
  formula is 4n + 4, but I also have to find the formula for the total
  number of infected cells, and despite trying many times to figure it
  out, I still have no clue.
  
  infected cells goes up by 4 cells each minute, i.e. 3 minutes is 13
  and 4 minutes is 25.
  
  I will be extremely appreciative if you help me with this.
  Thanks,
  Victor Lee
  ```
  
  ```
  Date: 10/23/2001 at 14:41:02
  From: Doctor Greenie
  Subject: Re: I cannot figure out the formula to a math virus
  
  Hi, Victor -
  
  This is a nice problem. There are many opportunities with this problem
  to be "clever" to make finding the answer quite easy. I will present
  the solution in exactly the way I found it, so that you can see how
  the effort required to reach a solution can be reduced by studying the
  problem to find the "easy" path.
  
  Let's start by drawing some pictures and making a table of our
  results....
  
  +---+      n=0
  |   |      new squares: 1
  +---+      total squares: 1
  
  +---+
  |   |
  +---+---+---+      n=1
  |   |   |   |      new squares: 4
  +---+---+---+      total squares: 5
  |   |
  +---+
  
  +---+
  |   |
  +---+---+---+
  |   |   |   |
  +---+---+---+---+---+      n=2
  |   |   |   |   |   |      new squares: 8
  +---+---+---+---+---+      total squares: 13
  |   |   |   |
  +---+---+---+
  |   |
  +---+
  
  +---+
  |   |
  +---+---+---+
  |   |   |   |
  +---+---+---+---+---+
  |   |   |   |   |   |
  +---+---+---+---+---+---+---+      n=3
  |   |   |   |   |   |   |   |      new squares: 12
  +---+---+---+---+---+---+---+      total squares: 25
  |   |   |   |   |   |
  +---+---+---+---+---+
  |   |   |   |
  +---+---+---+
  |   |
  +---+
  
  So we have
  
  figure number   additional squares   total squares
  --------------- -------------------- ---------------
  0                 1                  1
  1                 4                  5
  2                 8                 13
  3                12                 25
  ...
  
  The sequence we get for the numbers of squares for n = 1, 2, 3, ... is
  
  5, 13, 25, ...
  
  One way to find a formula for generating a sequence like this is the
  method of finite differences. Here is a link to a page in the Dr. Math
  archives where there is a lengthy discussion of this method:
  
  Method of Finite Differences
  http://mathforum.org/dr.math/problems/gillett.10.12.00.html
  
  While the method of finite differences is often very useful, I
  suspected I could find the answer more quickly by analyzing the
  figures and coming up with some shortcut for finding the total number
  of squares in the n-th figure.
  
  My approach (taken with too little forethought, as you will see later)
  was to break up the figure for the case n=3 as follows:
  
  +---+   +---+   +---+
  |   |   |   |   |   |
  +---+---+   +---+   +---+---+
  |   |   |   |   |   |   |   |
  +---+---+   +---+   +---+---+
  |   |
  +---+---+---+---+---+---+---+
  |   |   |   |   |   |   |   |
  +---+---+---+---+---+---+---+
  |   |
  +---+---+   +---+   +---+---+
  |   |   |   |   |   |   |   |
  +---+---+   +---+   +---+---+
  |   |   |   |   |   |
  +---+   +---+   +---+
  
  The number of squares in this figure is the number in the central
  "cross" figure plus the number in the four identical "stair-step"
  figures. The number in the central cross figure for the n-th case is
  (4n+1); the number in each of the four stair-step figures is the sum
  of the integers from 1 to (n-1), which I know to be n(n-1)/2.  So the
  total number of squares in the n-th figure is
  
  (4n+1) + 4(n(n-1)/2))
  = 4n + 1 + 2(n^2 - n)
  = 4n + 1 + 2n^2 - 2n
  = 2n^2 + 2n + 1
  
  When I got to this point, I tried to find a simpler way to write this
  formula; after staring at it for a moment, I realized that
  
  2n^2 + 2n + 1
  = (n^2) + (n^2 + 2n + 1)
  = n^2 + (n+1)^2
  
  In words, this formula is quite simple - the number of squares in the
  nth figure is the sum of the squares of n and (n+1).
  
  Checking this formula out quickly for n = 1, 2, and 3, we find
  
  n=1: n^2 + (n+1)^2 = 1+4 = 5
  n=2: n^2 + (n+1)^2 = 4+9 = 13
  n=3: n^2 + (n+1)^2 = 9+16 = 25
  
  So the formula appears to be correct, indicating I did not make any
  mistakes with my reasoning or with my algebra.
  
  But after I came up with this formula, I went back to look at my
  figures to see if I could come up with this same formula more quickly
  than the way I did. And just a bit of looking at the figure showed me
  what would have been a much quicker path to the formula.
  
  The quicker path I found was to break up the n-th figure in a
  different manner than I did originally. This time, I break up the n-th
  figure as follows:
  
  +---+
  |   |
  +---+---+---+
  |   |   |   |
  +---+---+---+---+---+
  |   |   |   |   |   |
  +---+---+---+---+---+
  
  +---+---+---+---+---+---+---+
  |   |   |   |   |   |   |   |
  +---+---+---+---+---+---+---+
  |   |   |   |   |   |
  +---+---+---+---+---+
  |   |   |   |
  +---+---+---+
  |   |
  +---+
  
  Now the figure is comprised of two parts. The number of squares in the
  first part is the sum of the first n odd integers, which I know to be
  n^2; and the number of squares in the second part is the sum of the
  first (n+1) odd integers, which I know to be (n+1)^2. So by breaking
  up the figure in this manner I am led immediately to the formula
  
  number of squares in n-th figure = n^2 + (n+1)^2
  
  Thanks for the interesting problem. I got some valuable problem-
  solving experience playing with it.
  
  I hope all this helps. Write back if you have any further questions on
  this.
  
  - Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/
  ```
  Associated Topics:
  High School Basic Algebra
  High School Number Theory
  High School Sequences, Series
  
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