If one number of a Pythagorean triplet is 5 then other two numbers are
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Answer:
Pythagorean triple (PT) can be defined as a set of three positive whole numbers that perfectly satisfy the Pythagorean theorem: a2 + b2 = c2.
This set of numbers are usually the three side lengths of a right triangle. Pythagorean triples are represented as: (a, b, c), where, a = one leg; b = another leg; and c = hypotenuse
There are only three types of Pythagorean triples that could exist given a number.
The first type is where the triple is a multiple of a smaller Pythagorean triple. For example, 6–8–10 is a multiple of 3–4–5. However, 5 is a prime number and so no such triples exist that include 5.
The second type is when the triple is a leg. As such, odd numbers and even numbers have two different methods of generating such a Pythagorean triple. With an even number, you square it, divide it by 4, and take one integer on each side of this value. However, 5 is not an even number so this is not of use. With an odd number, you square it, divide it by 2, and use the two closest numbers to that value. For example, with 7, you get 24 and 25. If you try this with 5, you get 12 and 13.
The third type is when the number given is the hypotenuse. As such, you reverse the processes described in the previous paragraph. To achieve one possible set, you get one leg by subtracting 2 (yielding 3 in this case) and you get the other by subtracting by 1 (getting 4), multiplying by 4 (getting 16), and taking the square root (which in this case yields 4). To achieve another possible set, you get one leg by subtracting 1 (getting 4), and then you get another by subtracting 0.5 (getting 4.5), multiplying by 2 (getting 9), and taking the square root (getting 3). In this case, both methods yield a 3–4–5 triangle.
Therefore, the only two triples that contain the number 5 are (5,12,13) and (3,4,5).
Below is some more fun for the people who are wondering why I wrote such a long answer:
You could use this analysis for any number. For example, given the number 28, you must first find Pythagorean triples containing all it’s multiples and then multiplying them. The triples include 7–24–25, which gets multiplied to become 28–96–100, and 3–4–5, which gets multiplied to become 21–28–35. Secondly, you find the triple with 28 as its leg. Square it (getting 784), divide it by 4 (getting 196), and get the two nearest integers (195 and 197). 28–195–197 is another multiple. Thirdly, you get the triples for which 28 is a hypotenuse. Neither methods creates an integer triple for which 28 is hypotenuse.
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Kurt Behnke, PhD Mathematics & Theoretical Physics, University of Hamburg (1981)
Answered December 13, 2017
Nice exercise. You don’t even have to use the hint that there are only two of them.
All Pythagorean triples are of the form (n2−m2,2nm,n2+m2)(n2−m2,2nm,n2+m2) for natural numbers n,mn,m.
Now the even number 2nm2nm cannot be equal to 5, and you end up searching solutions for one of the equations
n2−m2=5n2−m2=5 or n2+m2=5n2+m2=5
so 5 as a difference of two squares or a sum of two squares.
Let us start with the sum case, since this is a priori limited. The only squares smaller than 5 are 1 and 4, and by chance: 1+4=5 is part of the famous Pythagorean triple (3,4,5): 32+42=5232+42=52.
Now to the difference case: The difference between t
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