Side relations of triangles
How many non similar triangles exist such that the side-lengths of the triangle are positive integers and the areas of squares constructed on their sides are in arithmetic progression?
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1 solution
Exactly the family I found, albeit by inspecting numerical data in order to conjecture a pattern. Another infinite family of solutions is
a = 4 n 2 + 4 n − 1 , b = 4 n 2 + 8 n + 5 , c = 4 n 2 + 1 2 n + 7
Again, it's easy to show that for n ≥ 1 , a + b > c , b 2 − a 2 = c 2 − b 2 , and that g cd ( a , b , c ) = 1 , so no two of these triangles are similar.
How will u, v, b be coprime? Also, sir, how did you reach this approach?
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(1) If u , v , b had a common factor, then a , b , c would have a common factor. Since any triangle with integer sides is similar to one with integer sides that are coprime, it is OK for me to assume that a , b , c are coprime.
(2) The method for solving a 2 + c 2 = 2 b 2 is fairly standard.
If 1 ≤ a ≤ b ≤ c are coprime with a 2 + c 2 = 2 b 2 , then a , b , c must all be odd, and so we can define nonnegative integers u , v by u = 2 1 ( c − a ) , v = 2 1 ( c + a ) . But then u ≤ v and a = v − u , c = v + u , and b 2 = u 2 + v 2 , with u , v , b coprime.
If q > p > 0 and p , q are coprime, with 2 p q > q 2 − p 2 , we can have u = q 2 − p 2 , v = 2 p q and b = q 2 + p 2 . In particular, with q = n + 1 , p = n , we obtain a = 2 n 2 − 1 b = 2 n 2 + 2 n + 1 c = 2 n 2 + 4 n + 1 Provided that n ≥ 2 these values give us the sides of a triangle, since c ≤ a + b as well. All of these triangles are non similar, so there are infinitely many possibilities.