Find a solution for a^2+b^2+c^2+d^2+e^2=abcde in integers>0.

freemind

#1

Find a solution for a^2+b^2+c^2+d^2+e^2=abcde in integers>0.

**Posted:** Sun Mar 30, 2008 3:39 pm

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scorpius119

#2

an answer

how to arrive at it

**Posted:** Sun Mar 30, 2008 11:32 pm

campos

#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 Mr. Green

**Posted:** Tue Jun 03, 2008 6:35 am

Euclid's Turtle

#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?

**Posted:** Wed Jun 04, 2008 2:28 am

campos

#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...

**Posted:** Wed Jun 04, 2008 6:44 am