Problem Solving

Rahul@gurome wrote:

goyalsau wrote:You can apply this rule for any odd integer except 1

Every odd number is the a side of a Pythagorean triplet.
The b side of a Pythagorean triplet is simply (a^2 - 1) / 2.
The c side is b + 1.

You can apply this rule for any even integer except 2

Every even number is the a side of a Pythagorean triplet.
The b side of such a Pythagorean triplet is simply (a/2)^2 - 1.
The c side is b + 2.

Agree with Geva!
Let's try to understand how it is possible. That may help us to remember it.

Fact 1: The difference between the squares of two successive integers is always an odd integer.
Proof: Say the successive integers are n and (n + 1). Difference of their squares = (n + 1)² - n² = (2n + 1) => always an odd integer!
Use: Square of an odd integer is also an odd integer. Therefore, for any odd integer we can represent its square as the difference of the squares of two successive integers. In other words, we are getting a Pythagorean triplet with an odd number as one of them. And we can do this for any odd integer except 1. Because for 1, n = 0. Which results in the triplet (0, 1, 1). But we can't take that as Pythagorean triplet!

Let's check for some odd integers,

• (1) For 3, (2n + 1) = 9 => n = 4 => Other two sides are 4 and 5.
  (2) For 5, (2n + 1) = 25 => n = 12 => Other two sides are 12 and 13.
  (3) For 17, (2n + 1) = 289 => n = 144 => Other two sides are 144 and 145.

Same goes for even integers. Let's see,

Fact 2: The difference between the squares of two alternate integers is always an even integer.
Proof: Say the successive integers are n and (n + 2). Difference of their squares = (n + 2)² - n² = (4n + 4) = 4(n + 1) => always an even integer!
Use: Square of an even integer is also an even integer. Therefore, for any even integer we can represent its square as the difference of the squares of two alternate integers. In other words, we are getting a Pythagorean triplet with an even number as one of them. And we can do this for any even integer except 2. Because for 2, n = 0. Which results in the triplet (0, 1, 1). But we can't take that as Pythagorean triplet!

Let's check for some odd integers,

• (1) For 4, 4(n + 1) = 16 => n = 3 => Other two sides are 3 and 5.
  (2) For 10, 4(n + 1) = 100 => n = 24 => Other two sides are 24 and 26.
  (3) For 18, 4(n + 1) = 324 => n = 80 => Other two sides are 80 and 82.

Both of the above can be transformed to goyalsau's formulas. But I believe this explanation is easier than to remember those.

Geva@MasterGMAT wrote:I like you, man, but you're crazy . Ain't nothing about the above that is easy to either remember or apply.