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Numbers 1, 2, 3,...100 are written down on each of the cards A, B, C. One number is selected at random from each of the cards. One number is selected at random from each of the cards. Then find the probability that numbers so selected can be the measures of three sides of right angled triangles, no two of which are similar.

My Approach: One way to calculate number of combinations of three sides satisfying Pythagoras Theorem is manually there are 6 combinations (3, 4, 5), (5, 12,13) and so on. But if the number is large enough is there any mathematical way to find three numbers satisfying Pythagoras.

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    See [here](http://mathoverflow.net/questions/244491/is-there-any-formula-to-find-number-of-pythagorean-triplets-between-two-integers) if you want to find number of Pythagorean triplets between two integers. – 2017-02-20
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    What is the phrase "no two of which are similar" meant to restrict? – 2017-02-20
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    There are just $14$ primitive Pythagorean triples in $\{1,2,\ldots,99,100\}$ and just $50$ Pytagoran triples (primitive or not). – 2017-02-20

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There are not so many Pythagorean triples in $\{1,2,\ldots,99,100\}$. Primitive triples are given by $(u^2-v^2,2uv,u^2+v^2)$ with $\gcd(u,v)=1$ and at least one number between $u$ and $v$ being even - they are associated with rational points on the unit circle. On the other hand neither $u$ or $v$ can be larger than $9$ in our case, so the primitive triples of interest are given by $$ (u,v)\in\{(2,1),(4,1),(6,1),(8,1),(3,2),(5,2),(7,2),(4,3),(8,3),(5,4),(7,4),(6,5),(8,5),(7,6)\}$$ and they are $$ \underbrace{(3,4,5)}_{20},\underbrace{(15,8,17)}_{5},\underbrace{(35,12,37)}_{2},\underbrace{(63,16,65)}_{1},\underbrace{(5,12,13)}_{7},\underbrace{(21,20,29)}_{3}$$ $$ \underbrace{(45,28,53)}_{1},\underbrace{(7,24,25)}_{4},\underbrace{(55,48,73)}_{1},\underbrace{(9,40,41)}_{2},\underbrace{(33,56,65)}_{1},\underbrace{(11,60,61)}_{1}$$ $$\underbrace{(39,80,89)}_{1},\underbrace{(13,84,85)}_{1}$$ where the small numbers denote the primitive or non-primitive triples in $\{1,2,\ldots,99,100\}$ associated to them. It follows that there are only $50$ Pythagorean triples in $\{1,2,\ldots,99,100\}$, over $\binom{100}{3}$ possible triples.