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Post Posted: Mar 30, 2008, 6:39 am • # 1 


Determine a subset A\subset \mathbb{N}^* having 5 different elements, so that the sum of the squares of its elements equals their product.
Do not simply post the subset, show how you found it.

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Post Posted: Mar 30, 2008, 2:32 pm • # 2 


an answer

how to arrive at it
 
 
Post Posted: Jun 02, 2008, 9:35 pm • # 3 


i found (1,3,4,9,107) and (3,4,9,107,11555).

suppose we have a solution, and consider it as a quadratic in a. we have its discriminant equals (bcde)^2 - 4(b^2 + c^2 + d^2 + e^2). it has to be a perfect square so (bcde - k)(bcde + k) = 4(b^2 + c^2 + d^2 + e^2).

suppose that bcde - k = 2 and bcde + k = 2(b^2 + c^2 + d^2 + e^2). so, we have that bcde = b^2 + c^2 + d^2 + e^2 + 1.

consider it again as a quadratic in b. we have its discriminant equals (cde)^2 - 4(c^2 + d^2 + e^2 + 1). again we take
cde - k = 2 and cde + k = 2(c^2 + d^2 + e^2 + 1), so, cde = c^2 + d^2 + e^2 + 2.

consider it again as a quadratic in c. its discriminant equals d^2e^2 - 4(d^2 + e^2 + 2). take e = 3. we have to find a integer d such that
5d^2 - 44 = k^2. for d = 4 we have that 5\cdot 4^2 - 44 = 36.

if we return to previous equations, we find that 12c = c^2 + 27, so we find (c - 3)(c - 9) = 0... analogously we find (b - 1)(b - 107) = 0. if b = 1 we find (a - 107)(a - 1) = 0. if b = 107 we find (a - 1)(a - 11555) = 0, and that's how i found them :D
 
 
Post Posted: Jun 03, 2008, 5:28 pm • # 4 


campos, how do you get that discriminant equation? and why must it be a perfect square? To have 2 real roots, but in the same spot?
 
 
Post Posted: Jun 03, 2008, 9:44 pm • # 5 


the discriminant of the equation a^2-a\cdot bcde+(b^2+c^2+d^2+e^2)=0 equals b^2c^2d^2e^2-4(b^2+c^2+d^2+e^2)...

it has to be a perfect square in order for a to be an integer...
 
 
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