Question 464047
1. Suppose a,b,c,d are positive integers satisfying


*[tex \LARGE a^2 + b^2 = c^2 + d^2 = k]


Then,


*[tex \LARGE a^2 - c^2 = d^2 - b^2]


*[tex \LARGE (a-c)(a+c) = (d-b)(d+b)]


After this was some "educated trial-and-error." I simply thought of numbers that would factor in two different ways (e.g. 48 = 12*4 = 24*2). The only constraints were that the factors (e.g. 12,4 and 24,2) had to be both odd or both even, and distinct, so that a,b,c,d would be nonnegative integers. I obtained 15 = 15*1 = 5*3, which gives the equations


*[tex \LARGE a+c = 15]
*[tex \LARGE a-c = 1]
*[tex \LARGE d-b = 5]
*[tex \LARGE d+b = 1]


Solving this yields a = 8, c = 7, b = 1, d = 3, and it can be checked that


*[tex \LARGE 8^2 + 1^2 = 7^2 + 4^2 = 65]


which is the smallest such number (I hope).


2. Given


*[tex \LARGE a+b = ab] we can find b in terms of a,


*[tex \LARGE a = ab - b = b(a-1)]


*[tex \LARGE b = \frac{a}{a-1}]


This essentially defines a function b in terms of a, and there are infinitely many ordered pairs (a,b) that satisfy. For example, if a = 3, b = 3/2, and 3 + (3/2) = 3(3/2) = 9/2.