Question

Solve the following equation in natural numbers:

\(\large{x^2 + y^2 = 1997(x-y)}\)

Answer 1

@Abhishek619
@mathslover
@Miracrown
@mukushla
@hartnn

Answer 2

@ganeshie8

Answer 3

well, I think discriminant of quadratic must be a perfect square

Answer 4

one obvious solution is 0,0
:P

Answer 5

But that isn't Natural number..

Answer 6

Busted @hartnn

Answer 7

and this is just a quadratic
so you can use same method as for previous
as mukushla said,
discriminant = perfect square...


yeah, ok ...find all integer solution and eliminate those are not applicable, that my way of solving :P

Answer 8

See how big my heart is !!

Come on applaud and praise me.

Answer 9

I guess I would want to use a similar method to before. Express this as quadratic in "y" and then look for particular forms of the result that give us natural numbers. Let's try that
Guess I am a stickler for basic methods

Answer 10

hmm

Answer 11

And I am going to stick with mine @Miracrown ;)

Answer 12

170,145 ?
1827, 145 ?

Answer 13

sureeee do as you wish! :P

Answer 14

some obvious ones :)
(0,0) , (1997,0), (0,-1997)

Answer 15

Yeah sure . Sorry the only answers are:
(170,145); (1827,145); (1972,315)

Answer 16

oh nat numbers

Answer 17

\(t = 1997 \sqrt{ 1 - \cfrac{2}{x+y} } \)

Answer 18

you sure about 1972, 315 ?

Answer 19

i am not sure abt 1972

Answer 20

-4x^2 + 7988x + 1997^2
Okay, few notes
We need this number to be a perfect square
We also need it to be an odd perfect square

because we need -1997 = odd... odd+odd = even so 2 divides evenly the root

Answer 21

Yeah sorry I am going through 2 questions at a time and mixing up answers.

Answer 22

yup i got the answer

Answer 23

The only solutions are (170,145); (1827,145)

Answer 24

1972 and 315 is correct

Answer 25

no my answeer is wrong

Answer 26

lemme check it again

Answer 27

i got 170,145 as my answer

Answer 28

So the only thing that I didn't write is that we almost must demand -4x^2+7988x + 1997^2 > 1997^2 because natural numbers are positive
so
This really limits where we are on the parabola

Answer 29

@Miracrown You are correct in saying that.

Answer 30

--> -4x^2 + 7988 x > 0
---> 4x (-x + 1997) > 0. ---> 0 < x < 1997. Okay not that helpful >.<

Answer 31

Okay, the other constraints (-4x^2+7988x + 1997^2) = (2n+1) for some n
(-4x^2+7988x+1997^2) = m^2 for some m

Answer 32

We can do quadratic formula on these again and get further conditions on x... but just thinking if that's not too messy

Answer 33

I am going to make some simplifications
"b" = 1997
-4x^2 + 4 b x + b^2 -2n-1 = 0
x = -4b +- Sqrt[(4b)^2 - 4(-4)(b^2-2n-1)] all over -8

Answer 34

Do the pythagoras substitution.

Answer 35

Likewise with m^2

Answer 36

What do you mean there? o.O
you pythagorean triples

Answer 37

well this is better, using some algebra\[(x+y)^2+(1997-x+y)^2=1997^2\] rest will be easy @mathslover ha?

Answer 38

x = -4b +- Sqrt[(4b)^2-4(-4)(b^2-m^2)] all over -8

Answer 39

This is what I suggest :
for : a^2 + b^2 = c^2
(m > n)
a = (m^2 - n^2 )
b = 2mn
c = m^2 + n^2

Answer 40

continuing with my work... x = b/2 -+ 2 Sqrt[b^2 - 2n - 1]
Okay, this looks helpful. Because we want x to be a natural number. This means that we need b -+ 4 Sqrt[b^2-2n-1] to be even
but b = 1997, which is odd
odd+- even = odd
which is never divisible by 2

Answer 41

this means we demand that b^2-2n-1 = 0

Answer 42

Ugh, this means n = 1994004

Answer 43

Actually, it's just a contradiction, because this means that x can never be a natural number. So this means there's something a bit wrong here. Rethinking

Answer 44

a = x
b = y
c = \(\sqrt{1997(x-y)}\)

a = m^2 - n^2 = x
b = 2mn = y
c = m^2 + n^2 = \(\sqrt{1997 (x-y)}\)

Answer 45

yea not sure how to work with that
But I see the general idea of trying to fit the pythagorean triples to the answers

Answer 46

http://www.mathblog.dk/pythagorean-triplets/

I am reading this blog right now. May be, any of you will also want to look at this.

Answer 47

This is just reminds me very much of diophantine type equations and fermat's last theorem type problems. But with the squares, you can relate it to the triples, which are pretty well known. So it irks me a bit not solving it completely.
But yeah, I'll just get it later haha

Answer 48

Yep you are right with diophantine type equations here. http://eprints.usq.edu.au/20783/1/Yevdokimov_ASMJ_v25n1_PV.pdf

I have been doing some searches on google related to pythagorean triplet and found this. This will be solved by Pythagorean triplet, that I'm sure as we have to find another triplet satisfying the situation. But, am not sure how to do this.

Answer 49

Okay guys you did a lot of work now its my turn.

I am going to continue from where @mukushla left;

\((x+y)^2 + (1997-x+y)^2 = 1997^2\)

Answer 50

Now, since x and y are positive integers (or natural numbers), so,

I can write that;

\(0<x+y<1997\)

and \(0<1997-x+y<1997\).

Now I am putting \(a = x+y\) and \(b = 1997-x+y\). So, I only have to solve:

\(a^2 + b^2 = 1997^2\)

Thus, (a,b,1997) make Pythogorean triplets.

Now, since, 1997 is a prime number. So, \(gcd(a,b) = 1\) .

Now I am going to use Pythogorean substition, i.e. I am going to put,

\(a = 2mn\) and \(b = m^2 - n^2\). Also, I am asuming m>n. And I know that gcd(a,b) = 1.

Answer 51

Now I am going to us modulo,

Answer 52

just a quick question how do I write this in latex ?

Answer 53

\(\equiv\)

Answer 54

Thanks @rational.

Now I have,

\(m^2, n^2 \equiv 0,1,-1 (mod~5) ~\text{and}~ 1997 \equiv 2(mod~5)\)

So,

\(m,n \equiv \pm1~(mod~5)\)

Similarly,

Since,

\(m^2, n^2 \equiv 0,1~ (mod~3) ~\text{and}~ 1997 \equiv 2~(mod~3)\)

Thus,

\(m,n = 1,4,11,14~(mod~15)\).

Also, I assumen m>n. So,

\[\cfrac{ 1997 }{ 2 } \le m^2 \le 1997\]

Now \(\cfrac{1997}{2} = 998.5\) and \(\sqrt{998.5} =31.599\). Also, \(\sqrt{1997} = 44.68\). So, I only have to consider m = 34, 41, 44.

But, only m = 34 satisfy the previous conditions. So,

(m,n) = (34, 29)

So, (a,b) = (1972, 315), which leads to our answer.

Answer 55

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5

\(a = 2mn\) and \(b = m^2 - n^2\) (a and b can exchange themselves)

\(\color{blue}{\text{End of Quote}}\)

Answer 56

Not the best solution but it works.

Answer 57

@ganeshie8 , @hartnn , @mukushla , @Miracrown

Answer 58

While many school students are not taught about modulo, it really helps in easily solving very difficult problems.