Problem 313. Pythagorean perfect squares: find the square of the hypotenuse and the length of the other side

Created by Bruce Raine
Appears in Cody Problems in Japanese

Difficulty:Rate

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Given the square root of a square number, seed, and a range, n, find the square number, Z as well as the other side, y, the square root of a square number i.e. return the hypotenuse squared as well as the length of the other side. Note that n is the number of squares to search through starting with one.

HINT: Z = seed^2 + y^2 where Z = z^2, find Z first and then y.

Note that Z, seed^2 and y^2 are all perfect squares.

>> [z s] = findPerfectZ(3,6)

z = 25

s = 4

Solve

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Last Solution submitted on Dec 28, 2024

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Peter Wittenberg on 2 Jan 2014

There's a problem with the solution suite. For seed=12 and n=16, the proposed answer of 5, 12, 13 as a Pythagorean triple is indeed a good one. However, 9, 12, 15 is equally valid but not included as an answer. To avoid this, I would suggest changing the problem so that it requires finding the answer with the minimum Z^2 to avoid ambiguity.

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Problem 313. Pythagorean perfect squares: find the square of the hypotenuse and the length of the other side

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Problem 313. Pythagorean perfect squares: find the square of the hypotenuse and the length of the other side

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Last 200 Solutions

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Problem Recent Solvers56

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