Find the smallest number that can be expressed as the sum of two (nonzero) perfect squares in two different ways.
And what if the two perfect squares must be nonzero, positive, and different?
(In reply to answer
by K Sengupta)
Let N denote the required number.
Then, a^2+ b^2 = c^2+ d^2
(I) a, b, c and d are not all distinct.
Without loss of generality, we may take c=d = p, so that:
a^2+ b^2 = 2*p^2
Or, ((a+b)/2)^2 + ((a-b)/2)^2 = p^2....(#)
Since, the minimum possible Pythagorean Triplet is (3,4,5), we can substitute:((a+b)/2, (a-b)/2) = (4, 3); giving:
(a,b) = (7, 1), so that: p = 7 in terms of (#)
Consequently, the smallest number that can be expressed as the sum of two (nonzero) perfect squares in two different ways is 50.
(II) a, b, c and d are all distinct
N = a^2 + b^2 = c^2 + d^2
Wlog, let us assume that d>a and b>c.
Then, d^2 - a^2 = b^2 - c^2 (= r, say)
Or, r = (d+a)(d-a) = (b+c)(b-c)...(##)
Now each of (d+a, d-a) and (b+c, b-c) must possess the same parity. Otherwise, (##) would yield non integral values of (b,c) or (d, a) which is a contradiction.
Accordingly, r corresponds to the smallest number which is expressible as the product of two numbers with the same parity in precisely two different ways.
By inspection, we can easily see that r=15 = 15*1 = 5*3
Compating this with (##) yields:
(d+a, d-a, b+c, b-c) = (15,1,5,3)
Or, (a,b,c,d) = (7,4,1,8)
N = 7^2 + 4^2 = 1^2 + 8^2 = 65
Thus, smallest number that can be expressed as the sum of two (nonzero) and distinct perfect squares in two different ways is 65.