Anyone knows how to find a formula for......?

Question

If you are given a number
eg: 3
You'll straight away knows the other two which is 4 and 5.
eg: 40
You'll know the other two numbers are 399 and 401.

eg: 40
You'll know the other two numbers are 399 and 401.
It's something like Pythogoras Therom
you are given the smallest number, eg: 24, you'll straight away know the other two numbers which are 143 and 145.

And what if it is an odd number?
eg: 41

Answer 1

These are Pythagorean Triples. 3² + 4² = 5², and 40² + 399² = 401². Since you start with the smallest of the three, call it a, you want c² - b² = a². Since c² - b² factors, you have (c - b)(c + b) = a². So you work with factors of a². Note that if you add (c - b) and (c + b), you get 2c, and if you subtract them, you get 2b.

In the case of a = 3, a² = 9, and the factors of 9 pair up as just 1 • 9 and 3 • 3. Using 1 • 9, 1+9 = 10 = 2 • 5, so c = 5, and 9 - 1 = 8 = 2 • 4, so b = 4. You don't use the 3 • 3 pair because you'd get 0 as a factor.

In the case of a = 40, 40² = 1600 has LOTS of factor pairs. Note that 1 • 1600 leads to an ODD sum, and if a, b, c are integers, we need 2c to be even. But 2 • 800 works, with 2c = 802, c = 401, and 2b = 798, b = 399. Keep going through the factor pairs of 1600 and you'll find more Pythagorean triples: (40, 198, 202); (40, 96, 104); (40, 75, 85); and (40, 42, 58).

Answer 2

Are you sure that you're expressing this question correctly? *** (NO, you're NOT expressing it correctly --- see *** POSTSCRIPT, below!)

(3, 4, 5) is an obvious Pythagorean triple, for which it's true that :

3^2 + 4^2 = 5^2.

However, for such triples (those starting with an odd number) it's also true that they can be expressed as:

(2 n + 1, 2 (n^2 + 1), 2(n^2 + 1) + 1) --- the last two numbers differing by just 1. [n = 1 for the set (3, 4, 5).]

For such numbers then, with the first one, x, being odd, one can always get the other two by simply writing down (1/2)x^2 - 1/2 and (1/2)x^2 + 1/2, in fact. In other words, square x, divide by 2, and then simply subtract or add 1/2. THAT gives you an explicit specification of the other two numbers (given an odd smallest number, x), and therefore satisfies your criterion of being able to write the other two numbers down "straight away."

Unfortunately your second example doesn't fall into that category --- 40 is even, and the other two numbers differ by 2.
[LATER: I have more to say about this, below *** ]

Live long and prosper.

*** POSTSCRIPT *** I declare "FOUL" on this question. It was NOT PROPERLY POSED.

The way that you posed it said:

"eg: 3 You'll straight away know[s] the other two which is 4 and 5.
eg: 40 You'll know the other two numbers are 399 and 401."

BUT IN THE LATTER CASE YOU WON'T: While (40, 399, 401) is indeed a Pythagorean triple, by no means will you: "Know the others straight away" in the common meaning of that assertion, for the following reasons:

(a) I doubt that ANYONE keeps ALL such Pythagorean triplets in their heads, and it takes some WORK to find them. What's more, there may be many of them (see the next Answer); and

(b) There are for example several Pythagorean triplets starting with 40, e.g. (40, 399, 401); (40, 198, 202); (40, 96, 104); (40, 75, 85); and (40, 42, 58) --- see the next Answer. So even the mention of "40" coupled with the fact that it's the smallest number in a Pythagorean triple STILL isn't enough for you to know "straight away" that the other two numbers are 399 and 401. Only if you could provide an EXPLICIT FORMULA or FORMULAE that gives the numbers 399 and 401 AND ONLY THOSE NUMBERS would such a statement be justified. (Saying in words "It's the smallest number in a Pythagorean triple where the other two numbers are as close together as they can be" would not count as a "knowing straight away" specification.)

A SECOND FOUL is that you only added your extra challenge (in which you posit an intially odd number such as 41) 24 minutes after seeing my original answer, where I already gave you the general rule for numbers of that kind. I quote from my answer for odd initial numbers above, for the case of 41:

41^2 = 1681, so that the "other numbers" are 840 and 841 (i.e. half of 1681 +/- 1/2). What's more, these TWO numbers follow EXPLICITLY, unlike you own second example.

But shame on you!

Answer 3

I'm assuming you're trying to find all (X, Y, Z) such that X^2 = Y^2 + Z^2. If not, my apologies - wrong solution.

Let a and b be two integers, a>b.
Consider the three numbers:
X: a^2 + b^2,
Y: a^2 - b^2
Z: 2ab



If you square the three numbers, you get:
1. a^4 + 2a^2b^2 + b^4
2. a^4 - 2a^2b^2 + b^4
3. 4a^2b^2

Notice that the first = second + third.
From this we can see that we can generate an infinite number of triplets (X, Y, Z) by varying a and b and keeping X, Y, and Z all positive.

eg. a = 2, b= 1: X= 5, Y = 3, Z = 4
a = 3, b = 2: (13,5,12)
and so on.

Answer 4

I don't think there is enough information.
Give us more examples, or explain again how you 'know' that 40 becomes 399 and 401
More examples, please

[edit]

Ah, I see, pythagorean triples.

If your numer is even, you could use this simple formula to create your 2 numbers:

((your number / 2) ^ 2) + 1
((your number / 2) ^ 2) - 1

You are basically talking about a formula like this:
a^2 + b^2 = c^2
where
a = n^2 - m^2, b = 2nm, c = n^2 + m^2.
for integers m and n