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

the reciperocal pythagorean identity is the equation:

1/(a^2) + 1/(b^2) = 1/(c^2)

Try finding the smallest solution for positive integers, basically the smallest sum of the numbers a,b, and c.

No cheating! The answers of course can be found, but a rigourous method for finding it is the challenge. Any monkey could make a code or guess and check it.

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>>7587160

Please go away and do your homework yourself.

Really, it's not even that hard. Just literally check Wikipedia.

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>>7587170

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

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>>7587160

a=0

b=0

c=0

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>>7587176

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

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>>7587178

a=1

b=1

c=1

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>>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.

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

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>>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.

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>>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

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>>7587360

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

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>>7587360

nice

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>>7587360

Excellent job

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