I drew the book's diagram (p. 91) on my white board in Evans.  The  
proof can be expressed in terms of Fermat's method of descent.  If  
the square root of 1 + m^2 is a rational number, there is a right  
triangle with sides a and ma whose three sides are integers.  Take a  
as small as possible.  Write the length of the hypotenuse as ma + b,  
where b is a positive integer.  The sum of the two sides ma + a is  
bigger than ma + b, so b is a smaller integer than a.  The claim is  
that the right triangle whose two sides are b and mb has a hypotenuse  
that's also of integral length.  Because b is less than a, this  
contradicts the supposed minimality of a.

To prove the claim, note that (am)^2 + a^ 2 = (ma+b)^2.  This leads  
to the relation a^2 - b^2 = 2amb.  It follows that (mb)^2 + b^2 = (a-
mb)^2 by expanding out the square and doing a small amount of  
algebraic manipulation.  Thus the hypotenuse that we wanted to have  
integral length has length a-mb or mb-a, whichever is positive.  The  
diagram tells us that a is bigger than mb, but we don't really need  
to know this.

Ken R