 Quandaries and Queries Helloo I am a teacher who mainly works in special needs but I help out in the math department of a secondary school in England (children aged 12 - 16). I'm no mathematician but I enjoy math games and puzzles and I am a great fan of your site. The other day it occurred to some students that they could think of no square number which is an integer, which can be divided into two equal square numbers which are intergers, Or put another way, no squared integer when doubled can equal another square integer. For example 5 squared plus 5 squared is 50, but 50 is not a square number. There must be some way of demonstrating this but I suspect that it is more complicated than it looks - something to do with defining integers, maybe. I couldn't explain it - can you help? Thanks Mike Hi Mike, I received two answers to your question, one from Walter and another from Claude. Walter's answer contains some history of this question. I found it fascinating that you asked a question about numbers and both Walter and Claude talk about geometry. Harley It is an interesting problem. It has a classical geometric root as well. Consider an isosceles right angle triangle (half a square). If the two equal sides are integers - can the hypotenuse be an integer? By Pythagoris, you add the squares of the two equal sides, and see if the answer is also a square. [Well geometrically, there is a square on the hypotenuse, and it can all be constructed, say with ruler and compass.] The issue is whether this is the square with integer sides.] The answer is no and I will outline one proof below (Euclid's proof I believe). So the Greeks encountered this question - and discovered that it is not a square. Their conclusion was interesting. They could see the areas, the constructions etc., but not the numbers. Therefore, they concluded that geometry was reliable and precise - and numbers were a source of trouble. Only trust numbers when they have a geometric basis. We tend to take the reverse position. Numbers are exact, geometry, particularly geometry with figures, is less exact, less reliable. Therefore, only trust geometry when it can be put into numbers! A complete shift in beliefs and attitudes. So for a proof. The problem starts with a2 + a2 =? b2 or 2 a2 = b2. Assume there is a solution, and that a and b are the smallest possible whole number solution. Therefore, a and b have no common factor. [If they did,we would divide and get a smaller number.] b2 must be an even number - so b must be divisible by 2. With b= 2c, we have 2 a2 = 4c2 or a2=2c2, This new pair (c,a) is a smaller solution than (a,b). But we assumed, (a,b) was the smallest solution - a contradiction! Put more strongly root(2) = b/a does not have a rational solution (a, b whole numbers). So root(2) is irrational. Finding geometric objects which were not ratios of other whole numbers was disturbing to the Greek mathematicians. If you believe in the music of the spheres - and ratios as the core of beautiful patterns, these are monsters! Walter There is a way to construct squares which are "almost" as you ask: the second is two times the first, plus or minus one: Start with s = 1, d = 1 ("s" for single, "d" for double), and note that 2 s2 = 2 and d2 = 1, a difference of 1. Now proceed as follows to find a new s and a new d: The new s will be the old s plus the old d, the new d will be the old d plus two times the old s. In this way, (s,d) changes from (1,1) to (2,3). Do it again and (2,3) changes to (5,7), and again to get (12,17) ... In this sequence, we always have 2 s2 - d2 equals plus or minus 1; if you are willing to settle for this little difference, you get infinitely many solutions. Now, enough with the masquerade: "s" really stands for side, and "d" for diagonal: the diagonal of a square of side 5 is approximately 7, the diagonal of a square of side 12 is approximately 17, and so on. More generally, the diagonal of a square of side s is root(2) s. If you find numbers s and d such that 2 s2 = d2, then  d/s is root(2). I bet you already know that this is impossible. (That's Walter's proof above.) Claude Go to Math Central