Question 75320
Here's one approach:
Let n be the multiplier. The first multiple of 5 is 5n, the next consecutive multiple of 5 is 5n+5, and the third consecutive multiple of 5 is 5n+10.
From the problem description, you can write:
{{{(5n)^2+(5n+5)^2 = (5n+10)^2+125}}} Simplify and solve for n.
{{{25n^2+25n^2+50n+25 = 25n^2+100n+100+125}}}
{{{25n^2-50n-200 = 0}}} Factor out 25.
{{{25(n^2-2n-8) = 0}}} So...
{{{n^2-2n-8 = 0}}} Factor.
{{{(n-4)(n+2) = 0}}} Apply the zero product principle.
{{{n = 4}}} and/or {{{n = -2}}} 
So, you actually get two multipliers.  Let's check both of them:
n = 4
{{{(5*4)^2+(5*4+5)^2 = (5*4+10)^2+125}}} Simplify.
{{{20^2+25^2 = 30^2+125}}}
{{{400+625 = 900+125}}}
{{{1025 = 1025}}} This checks so the three consecutive multiple of 5 are:
20, 25, and 30

Now let's try n = -2
{{{(5(-2))^2+(5(-2)+5)^2 = (5(-2)+10)^2+125}}} Simplify.
{{{(-10)^2+(-5)^2 = (-10+10)^2+125}}}
{{{100+25 = 0+125}}}
{{{125 = 125}}} This also checks so could the three consecutive multiples of 5 also be?
-10, -5, 0 Well...not really because 0 is not a multiple of 5, is it?