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 Square challenge (Posted on 2004-01-20)
Find the smallest number that can be expressed as the sum of two (nonzero) perfect squares in two different ways.
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And what if the two perfect squares must be nonzero, positive, and different?

 See The Solution Submitted by SilverKnight Rating: 2.0000 (2 votes)

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 Problem Solution With Explanation | Comment 14 of 16 |

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)
This yields:
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.

 Posted by K Sengupta on 2007-05-19 04:35:01

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