$\textbf{ My Solution : } $ We will take two cases here, Case 1 : Either one of $a$ or $b$ is 30 WLOG, let $ a = 30 \hspace{4mm} \implies 30^2 = c^2 - b^2 \implies 30^2 = 2^2 \cdot 3^2 \cdot 5^2 = ( c -b)(c+b) $ here, both of $ ( c-b) $ and $ (c+b) $ must be even Since, $ (c - b) < (c+b) $ We will consider the cases Sub-Case 1: $ (c - b) = 2 $ and $ (c + b) = 2 \cdot 3^2 \cdot 5^2 $ $ \hspace{3mm} \implies c = 226 , b = 224 $ and $ a = 30 $ works Similarly for $c = 226 , b = 30 $ and $ a = 224 $ work Too! Sub-Case 2: $ (c - b) = 2 \cdot 3 $ and $ (c + b) = 2 \cdot 3 \cdot 5^2 $ $ \hspace{3mm} \implies c = 78 , b = 72 $ and $ a = 30 $ works Similarly for $c = 78 , b = 30 $ and $ a = 72 $ work Too! Sub-Case 3: $ (c - b) = 2 \cdot 3^2 $ and $ (c + b) = 2 \cdot 5^2 $ $ \hspace{3mm} \implies c = 34 , b = 16 $ and $ a = 30 $ works Similarly for $c = 34 , b = 30 $ and $ a = 16 $ work Too! Sub-Case 4: $ (c - b) = 2 \cdot 5 $ and $ (c + b) = 2 \cdot 3^2 \cdot 5 $ $ \hspace{3mm} \implies c = 50 , b = 40 $ and $ a = 30 $ works Similarly for $c = 50 , b = 30 $ and $ a = 40 $ work Too! Now, Count total solutions for this Case Case 2 : When $ c= 30 \implies a^2 + b^2 = 30^2 $ here we find $ a = 24 , b = 18 $ and $ a=18 , b=24$ as Only Solutions Hence, $ ( a , b , c) = ( 30 , 224 , 226 ) , ( 224, 30 , 226 ) ,(30,72,78),(72,30,78),(30,16,34),(16,30,34),(30,40,50),(40,30,50),(24,18,30),(18,24,30) $