The side lengths of a right triangle are each an integral number of units. If one of the legs is 13 units, what is the perimeter of the triangle?

Thanks
~Zach

All Pythagorean Triples of the form x^2 + y^2 = z^2 derive from x = k(m^2 - n^2), y = k(2mn), and z = k(m^2 + n^2) where k is any positive integer and m and n are arbitrary positive integers, m greater than n. Pythagorean Triples that have no common factor, or a greatest common divisor of 1, are called primitive. Those with a common factor other than 1 are called non-primitive triples. Primitive Pythagorean Triples are obtained only when k = 1, m and n are of opposite parity (one odd one even) and have no common factor, and m is greater than n. (For x, y, & z to be a primitive solution, m and n cannot have common factors and cannot both be even or odd. Violation of these two limitations will produce non-primitive Pythagorean Triples.)

Assuming k = 1, and that your statement "If one of the legs is 13 units" means one of the legs at 90º to one another is 13 units, then 13 is the side which derives from m^2 - n^2. Quick observation shows that m must be 7 and n must be 6 making 13 = 7^2 - 6^2. With m = 7 and n = 6, the three sides become a = 7^2 - 6^2 = 13, b = 2(7)6 = 84 and c = 7^2 + 6^2 = 85.

Since the perfect squares are the sequential sum of the odd integers starting with 1, the sum of the 2nd and 3rd squares, 4 + 9 = 13 deriving from m = 3 and n = 2 which leads to another Pythagorean triangle where a = 3^2 - 2^2 = 5, b = 2(3)2 = 12 and c = 3^2 + 2^2 = 13.

No values of m an n can yield a side of 2mn as it always results in an even number.

To find the perimeter of a triangle, you need to know the lengths of all three sides. In this case, we are given that one of the legs of the right triangle is 13 units. However, we still need to find the lengths of the other two sides.

Since this is a right triangle, we can use the Pythagorean theorem to find the lengths of the other sides. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Let's denote the length of the other leg as x and the hypotenuse as c. According to the Pythagorean theorem, we have the equation:

x^2 + 13^2 = c^2

Now, we can solve for x and c:

x^2 + 169 = c^2

To find integral solutions for x and c, we can examine the relationship between the squares of numbers. We know that if a number squared ends in 9, then the number must end in either 3 or 7.

Therefore, we can try various values of x that end in 3 or 7, and check if there is an integer value for c such that x^2 + 169 = c^2.

Let's try some values:
x = 3
3^2 + 169 = 178
This is not a perfect square.

x = 7
7^2 + 169 = 250
This is not a perfect square either.

Based on this pattern, it seems that the value of x will need to be larger than 7. We can continue trying different values of x until we find a perfect square for c^2.

x = 9
9^2 + 169 = 250
Still not a perfect square.

x = 12
12^2 + 169 = 325
Still not a perfect square.

x = 15
15^2 + 169 = 394
Still not a perfect square.

x = 16
16^2 + 169 = 433
Still not a perfect square.

x = 20
20^2 + 169 = 569
Still not a perfect square.

x = 25
25^2 + 169 = 744
Still not a perfect square.

x = 26
26^2 + 169 = 845
Still not a perfect square.

At this point, we can conclude that there are no integral solutions for x and c. Therefore, we cannot determine the lengths of the other two sides of the right triangle with integral values.

Without knowing the lengths of all three sides of the triangle, we cannot calculate its perimeter.

I apologize for the inconclusive answer. If you have any other questions, feel free to ask.

Look for three integers that satisfy

a^2 + b^2 = c^2

Have you tried 5, 12, 13? That is the only possibility with a 13-unit side.

Add them up for the perimeter.