Hello my friends. I have a problem for you today. the reciperocal

>>7587160
Please go away and do your homework yourself.
Really, it's not even that hard. Just literally check Wikipedia.

>>7587170
This is certainly not my homework, it was a problem for a math competition I supervised.

>>7587160
a=0
b=0
c=0

>>7587176
I am sorry but 0 is not a positive integer nor is this a solution set for this equation.

>>7587178
a=1
b=1
c=1

>>7587179
If you plug that into the equation, you will get 2 = 1

I will give a hint and say that a set of valid (a,b) values must follow
sqroot(a^2 + b^2) = z
Where z is some positive integer.

1/15^2+1/20^2=1/12^2

>>7587194
Congratulations, but as I said in the OP, this is not the challenge of the question. A general solution is what is difficult to find, finding the first solution is just a question to guide you.

>>7587210
a^-2+b^-2 = c^-2
is equivalent to
ab/sqrt(a^2+b^2) = c
since sqrt(a^2+b^2) is an integer,
a,b,sqrt(a^2+b^2) is a pythagorean triplet
so we need to find a pythagorean triplet (x,y,z) that satisfies
xy | z
we can get that by just taking any triplet (x,y,z) and multiplying it by z, since
(xz)^2+(yz)^2 = z^4
still holds and
(xzyz)/z^2=xy

the smallest pythagorean triplet is 3^2+4^2=5^2
so the smallest solution to your problem is
a = 3*5, b=4*5, c=5*5

>>7587360
oops, i meant c = (3*5)(4*5)/(5*5)=12

>>7587360
nice

>>7587360
Excellent job