Things We Learned on Square Day

1. Yesterday was squareday - for everybody, but it meant more to me. Yesterday was 4/25/09 – all square numbers. In a given century this'll happen 135 days out of 36,525 about 1 in 250 days - so not that big of a deal. But for me it marked my 36^th birthday – also a square number. This was the only day in my life when this will happen. In fact, for around 1 in 400 people this will only happen once. For everyone else it will never happen. Of course, if you were born on one of the 15 square-days in '00 you should get at nine of these, maybe ten if you don't smoke!
   
2. Oh, your birthday is all primes! Congrats. When will the special birthday come for you when your age and the year are both primes? Never. Sorry, if you were born in an odd-numbered year, you'll turn an even age in odd years (primes must be odd), and an odd (potentially prime) age in even years.
   
3. What do you call the special class of numbers that individually are the difference between two consecutive square numbers? Let's see. Let the first consecutive number be x and the second (x + 1). The difference in the squares is (x + 1)^2 – x^2 = y. Where y is one of the special class of numbers. Well, if you remember your FOIL method, y = 2x + 1. In order to ensure that both x and y are integers, the only stipulation is that y be an odd number. That's it. All odd numbers can be created by the difference of two consecutive squares.
   
4. Trying to find a square-root without a calculator? Check this out this page. http://www.homeschoolmath.net/teaching/square-root-algorithm.php
   
5. So, some numbers have integer square roots (4, 25, 9, 36) but most have non-integer irrational number square roots. Is it possible to have a non-integer rational square root? I don't think so. Coming back to the FOIL method, we could write an equation like: (a + x/y)* (a + x/y) = P. All of which are integers. Where a^2 < P (P is the number we're taking the square root of); and x<y. A little manipulation gets us to:
   

   (P – a^2) = 2x/y + x^2/y^2
   with a little more manipulation we can get to 2xy + x^2 = Ky^2,
   

   Where K is some integer multiplier.
   Solving for x: x = y±√(4*y^2+4*K*y^2) = y ± 2*y*√(2*K)
   

   x = y(1 + 2√2K).
   

   Since √2 (square-root of 2) is irrational, no integer values of x,y or K will produce a rational number.