First, suppose n is the shortest side of the triangle, m, k are other two sides. According to Pythagorean Theorem, we know *n*^{2} + *m*^{2} = *k*^{2}

just do a change *k*^{2} - *m*^{2} = *n*^{2}

futherly (*k* + *m*)(*k* - *m*) = *n*^{2}

as we know, *n*^{2} × 1 = *n*^{2} so we can suppose that *k* + *m* = *n*^{2}, *k* - *m* = 1 [because we have SPJ]

easily, we get

because the side is a interger, so this solution can only be used when n is a odd.

So how to deal with even? we find that if (k-m) is odd, the solution is suitable for odd. On the other hand, we guess that if(k-m) is even, the solution is suitable for even.

just as this,

so, we get

this is an *O*(1) algorithm.

So if you want to see more clearly, you can see the formual on this page. if you have some questions, i am willing to help you.:) [my english is not very good, so please pass some grammer fault.:P]

Auto comment: topic has been updated by I_am_SB (previous revision, new revision, compare).why did you remove the anime background solution. This is how all solutions should be presented.

well, at first, i thought CF don't support LaTex. So i added this picuture to let others see the formula But a very kind buddy told me how to print LaTex. therefore, i thought it's no necessary to keep this picuture on this solution:P

How do you prove that n is a cathetus of any right triangle ?

We don't need to mind that because we have a special judge. suppose it is a short side is much easier to solve.

Thank you for your explanation! It's really helpful.

I have a question.

This solution indirectly imply that every hypotenuse of some valid triple is always cathetus (leg) of other valid triple (since by definition of the task they would give you cathetus (leg) as well as hypotenuse). If it were not true your formulas would not work. But your formulas work well, so it's true.

How do you prove this? (every hypotenuse of some valid triple is always cathetus (leg) of other valid triple)

We don't need to mind that because we have a special judge. suppose it is a short side is much easier to solve

What do you mean special judge? The problem states that:

As I can see why your solution works: if you given hypotenuse

csuch thata^{2}+b^{2}=c^{2}form Pythagorean triple, then there is another valid Pythagorean triple such thatc^{2}+x^{2}=z^{2}and therefore you calculate other sidesx,zfromcas cathetus (leg).The question is why if

a^{2}+b^{2}=c^{2}is Pythagorean triple then exists another Pythagorean triple such thatc^{2}+x^{2}=z^{2}?you know, there are multiple answers, right?So the case that n is a short side is exist.

Yes, there are multiple answers but my question is different.

Suppose you are given in some test case

nas hypotenuse of some valid triplea^{2}+b^{2}=n^{2}.If there are no othervalid triple wherenis cathetus (leg) such thatn^{2}+y^{2}=z^{2}then your solutionwould not workbecause your formulas work only ifnis cathetus.But in practice, I see for every valid triple

a^{2}+b^{2}=n^{2}there is another valid triple such thatn^{2}+y^{2}=z^{2}.For example, let's take triple (8, 15, 17), 17 is hypotenuse here. Luckily, there another valid triple where 17 is cathetus (17, 144, 145). In practice for every valid (

a,b,c) there exists (c,x,z).But why? I don't understand. I thought about some number

cwhich only exists as hypotenusein some triple (a,b,c). In turns out that this number doesn't exist.You can't just assume that you can interpret any

nas cathetus. You have to prove that it's actually true.May be you didn't take into account this fact and just luckily gets accepted.

I want mathematical proof of this fact.

there must be such case, because any n(n>2) can find m, k that n^2 + m^2 = k^2

Amazing! I didn't know that. Thanks!

I thought there are "holes" where for some

nyou can't findm,ksuch thatn^{2}+m^{2}=k^{2}.It turns out your formulas can be used to prove by induction (base case 3) that for any

nthere ism,ksuch thatn^{2}+m^{2}=k^{2}.UPD: I was confused by this article: formulas for generating Pythagorean triples. I thought that if you need such complicated methods for generating Pythagorean triples, there should be numbers for which Pythagorean triples doesn't exist.

UPD2: I see what this wiki article is talking about. The article discuss methods to generate all possible valid triples. For example, all of these triples are valid (9, 12, 15), (9, 40, 41), (12, 35, 37) but your method generate only (9, 40, 41) and (12, 35, 37) but not (9, 12, 15). While methods discussed in this article allow to generate triples such as (9, 12, 15). I.e. your method is enough exactly what task 707C require (

find any valid triple forn).Actually all those mathematical formulas from the top of the post ARE proof that any n>2 can be used as a cathetus in a pythagorean triangle. That's the whole point of the post.

The formulas prove

everyn is cathetus of a valid triple, so the valid hypotenuses definitely are too, right?it doesn't matter whether n is cathetus of a valid triple. n can be short side or hypotenuse, but suppose it is a short side is eaiser for us to solve.

I clarify my question. Please, see my comment above.

You missed my point. His formulas show that you can find a triangle with n as cathetus for

any integern≥ 3.So clearly there isn't an integer

a^{2}+b^{2}=n^{2}that isn't a cathetus simply because there isn't any integer at all that isn't a cathetus.but we choose n as the shortest side is not wrong.

Thanks! Now, I understand.

If this problem's input output be like that how to solve it ?

InputEach case will contain one positive integer c (1 ≤ c ≤ 500,000), the length of the hypotenuse of the right triangle.

OutputFor each case, print the case number first, then print the other two sides a and b. If no such a, b can be found which satisfies the above condition print “Impossible” without quotes. If there are multiple pair of (a, b) exists, print the one with smallest possible a.

5

Case #1: 3 4

6

Case #2: Impossible

25

Case #3: 7 24

"because the side is a interger, so this solution can only be used when n is a odd." but why? i dont understand it.can u explain pls?

when n is an odd, (n^2+1)%2=0, so that the side is an interger. just reverse that.

Hi! I got the point that there isn't an integer that can't be made into the cathetus. But what happens if n= 10^9 ? If you consider it to be cathetus then either of m or k will be hypotenuse (larger than 10^9 ) hence it will violate problem constraint(1 ≤ m, k ≤ 10^18) .

Thanks in advance :)

-Tachibana_Kanade- thank you very much but i had another view

for any triangle pythegorean

a =m^2 — n^2 b = 2 *m *n c= m^2 + n^2

## where m>n and one of them is odd and one of them is even and gcd(n,m)=1

so a is always odd and b is always even . given x . if x is even then x =b x =2 *m *n now i should assume m and n to satisfy the condition n=1 odd m =x/2 even because x>=4 even number and gcd(1,m)=1 true for condition. so we get a and b .. and if x is odd it will be equal a.. is my approach right for now ?

I am not quite sure coz i don't get u clearly :(

would u like to be more specific?

If this problem's input output be like that how to solve it ?

InputEach case will contain one positive integer c (1 ≤ c ≤ 500,000), the length of the hypotenuse of the right triangle.

OutputFor each case, print the case number first, then print the other two sides a and b. If no such a, b can be found which satisfies the above condition print “Impossible” without quotes. If there are multiple pair of (a, b) exists, print the one with smallest possible a.

5

Case #1: 3 4

6

Case #2: Impossible

25

Case #3: 7 24

A case where this could break is even numbers whose factor is only 2 i.e. powers of (2,4,8,16)

Could you suggest a solution for that?

can you please say how did you get k+m = n^2 and k-m = 1 and k = (n^2 + 1) / 2 and m = (n^2 — 1) / 2 ??

"because we have SPJ" wat ?

He only meant Special judge. To be more specific, the judge is accepting any valid triplet. Like suppose you are given a number x as an input so (a,b,x) and (x,y,z) both are acceptable. But the point of the whole proof is to prove that for any valid triplet (a,b,c) where c is acting as an hypotenuse there exists at least one triplet where this c acts as a leg(smallest) side which is of the form of (c,d,e).