Number Theory in Mathlink docx

Namdung 5 Some NT'ists know this, but I couldn't even guess the right 43th DeMO German Mathematics Olympiad Problem 431146 Given a Cartesian coordinate system in the plane, a point with

Number Theory in Mathlink

Notice: There’re some mistakes in some solutions, so you have to read it carefully 1) system of equation including a prime number

Solve the following system of equations for all primes p with integer solutions x,y ! (1) p + 1 = 2x^2

(2) p^2 + 1 = 2y^2

Hint

1) OK, this is a German question

You can look up the solution in the 1997 - 1998 book,

http://www.unl.edu/amc/a-activities/a4-for-students/mc97-98-01feb.pdf

at page 41

2) it also appears in Po Leung Kuk invitational math competition(? or something like that, just for junior form, i.e grade 7-9) 5 years ago i think

2) diofant1

Solve the equation in Z:

x^4+6x 2 y 2 +y^4=z 2

Solution

(x 2 +y 2 ) 2 + (2xy) 2 = z 2 (given)

(x 2 +y 2 ) 2 -(2xy) 2 = (x 2 -y 2 ) 2

denote a=x 2 +y 2 and b=2xy

the above two equations imply

a^4 - b^4 is a perfect square

we will prove that this is not possible

assume there xists a nonempty set S s.t when a \varepsilon S ,then there xist natural numbers b and c satisfying a^4 - b^4 = c^2.by well ordering there is a least element ,suppose that element is a which implies

gcd (a,b,c) = gcd(a,b)=gcd(b,c)=gcd(c,a) = 1

consider cases

CASE 1 (c is even)

=>a and b are odd and 4lc 2

gcd(a,b)=1 and gcd(a 2 +b 2 ,a 2 -b 2)=2

let x= \sqrt [(a 2 +b 2)/2] and y = \sqrt [(a 2 -b 2)/2] then x and y are both integers and x^4-y^4=(ab) 2 but a>b implies x<a

which is a contradiction

CASE 2 (c is odd)

=> wlog ,a is odd and b is even now use pythagorean substitution and there xist integers

r and s such that gcd(r,s)=1 and a 2 =r 2+s 2 ,c 2 =r 2-s 2 and b 2 =2rs

xactly one of r and s is even.assume r' to be even and s' to be odd

r'/2=x 2 and s'=y 2 then

a 2 =4x^4+y^4

again by pythagoras,there xist integer m and s.t gcd (m,n)=1 and

a=m 2 +n 2 , 2x 2 =2mn and

y 2 =m 2 -n 2 since x 2 =mn and gcd(m,n)=1 m=p 2 and n=q 2

.so y 2 = p^4-q^4.thus (p,q,y) is a new solution with p= \sqrt m \leq x= \sqrt (r'/2)<r' \leq a,which is a contradiction

thus there do not xist solutions to this equation

• Nice soln Galois! This a well known problem: there are no numbers x and y s.t x^2+y^2=z^2 and x^2-y^2=t^2 This was used before on this forum I think I don't remember for which problem though Bye!

• There are many simmilar problems regarding this one which are done using galois

method

1)Prove that x^2+xy+y^2=36^2 has no solutions in Natural numbers \{0}

2)Solve in N* the equation 3x^4+10x^2y^2+3y^2=z^2

cheers!

3) Solve in integers :

(1) xy = y2001

(2) x2 + 15y2 = 22000

Solution

For the first one:

If x=1 then y=1 If x=-1 then y=-1 x=0=>y=0=>x^y=0^0 contradiction (same if y=0), so x<>0 Let's take the case |x|>1 It's obvious that y>=-1 because if y<=-2 then the left term isn't an integer while the right term is If y=-1 then x=-1 (we already have that soln) This leaves us with y>0

y=p1^a1* *pk^ak where pi are primes x must have the same prime divisors, so

x=p1^b1* *pk^bk (ai,bi>=1) The equation is translated to y*bi=2001*ai for all i, or p1^a1* *pk^ak*bi=2001*ai for all i from 1 to k (*) Let's assume one of pi doesn't divide

2001 (let's assume it's p1) This means that p1^a1|a1, which is impossible because for any p1>=2 p1^a1>a1 This means that p1|2001 It's the same for all the other primes, so k=1,

2 or 3, so y=3^a1*23^a2*29^a3 or 3^a1*23^a2 or any other case that might come out of this Let's consider the case when equations are 3^a1*23^a2*29^a3*bi=3*23*29*ai for all i This is equivalent to 3^(a1-1)*23^(a2-1)*29^(a3-1)*bi=ai for all i It's pretty

obvious that the left terms are greater than the right part if ai>1 for some i, so all ai must

be <=1 This means that all possible values of y are the divisors of 2001: 3, 23, 29, 3*23 etc

Conversely, it's easy to see that all divisors of 2001 are suitable values of y For example,

if y=3*23, then x=(3*23)^29 and we're done; if y=29 then x=29^(3*23) and so on Is it right, Arne? I might have missed sth

4) vietnam 2003

a problem from vietnam 2003

Find the largest positive integer n such that the following equations have integer solutions in x, y_1, y_2, , y_n:

(x + 1)^2 + y_1^2 = (x + 2)^2 + y_2^2 = = (x + n)^2 + y_n^2

Solution

Lemma: If u^2 + v^2 + 2 = 2w^2 then u, v are even and w is odd

Proof: Clearly u, v either both even, either both odd If u, v odd then LHS is divided by 4 but not by 8, that is impossible, because if RHS divided by 4 then it divided by 8 If u, v are even then LHS not divided by 4 => w is odd

If n>3, we have y_1^2 + y_3^2 + 2 = 2y_2^2 (why?)

y_2^2 + y_4^2 + 2 = 2y_3^2

From lemma we have y_2 odd and even at the same time, contradiction

For n=3 x=-2, y_1=0, y_2=1, y_3=0 fulfil

Namdung

5) Some NT'ists know this, but I couldn't even guess the right

43th DeMO (German Mathematics Olympiad) Problem 431146

Given a Cartesian coordinate system in the plane, a point with coordinates (x; y) is called

a grid point if and only if his coordinates x and y are integers Does there exist a circle with exactly five grid points lying on its periphery?

Solution

Nice solving for a German problem!!! :)

This is a problem that needs some spirit of observation ;]

There are two ideas that may bring you to its solution:

1) The equation sin^2(x)+cos^2(x)=1 has only four integer solutions;

2) If a grid point lies on the periphery then its symmetric for the center of the circle also lies on the periphery of the circle

1)One knows that the parametrical equations of the circle are :

x=a+Rcos(x)

y=b+Rsin(x), where (a,b) are the coordinates of the center of the circle, and R is its radius

But cos(x) and sin(x) are integer only for four points on the circle, so there exist only 4 grid points on the periphery of the circle (this may be difficult to understand for R irrational, but when one looks at the circle and at the coordinates, everything becomes clear )

2) This is a solution only for circles with grid center points :

Let M(c,d) be a point on the periphery of the circle, then its symetrical point M'

(according to the center of the circle) is also grid, because its coordinates are (2a-c,2b-d)

So there can be only an even number of grid points on the periphery of the circle, and because 5 is even there doesn't exist a circle with such a property

This solution came to my mind quickly, but its generality is hard to prove

• The complete solution is quite straightforward once you have the explicite

answer: The circle with equation passes through the points (0; -8), (5; 2), (-5; 2), (6; 0) and (-6; 0) and through no other lattice points About the general theorem, I think there is some information on the Mathworld site

6) nice

find all integer values that

((a + 1) / b) + ((b + 1) / a)

can assume and determine all pairs of integers a,b, giving these solutions

Hint

Nice, but classical If I'm not wrong, infinite descending shows that the only integer valus are 3 and 4 And if we plug these values, we will find to Pell equations, which aren't hard to be solved

7) Seem to be easy.

Was about to be a test for 10 grade student

Does it exist integer numbers satisfying integer,we can find at least

one number so that

Disscard

1) x also has to be an integer, right?

2) Wait I must be misreading this

Couldn't you just say that x=1, then find three integers that add up to n!?

3) Not really : you have to find a fixed triple which satisfies this for any After you've found such a triple you're only allowed to move

4) Hi, some questions:

i)Find all triple (a.b.c) in such that:there exits satisfied for any n>= then exits an integer x satisfied:

in part i) I only prove c=0 if ab<>0

ii)If we change to a polynomial in Z[x] whose degree >2 then what

is the conclusion of the problem?

5) I think I've reduced it to the question: do there exist s.t is a perfect square for all ?

Solution

there are no such (a,b,c): just use the condition for n=1,2,3,4 (you don't need more) and use the fast for any polynomial P with integer coefficients let

wlog we have (else translate x) and (actually, is only possible for -1 and 1 by the fact, thus by sending -x to x we can assume ) then we have some x such that it follows that

and , hence or thus, we have in the first case , and these equations lead to , thus , hence , contradiction

now take the condition for n=4: the discriminant must be a square, contradiction

• Very nice and good solution.But what about

8) Table with infinite number of fields

Email

A table with an infinite number of fields contains numbers.Empty fields are considered to contain zeros.In any square with sides along the grid lines is the absolute value of the sum of the numbers equal or less 1.Regarding these conditions, determine the maximum value of the sum of all elements in a rectangle 5 x 1

The problem is called "Der Burische Hammer" after Wolfgang Burmeister

Discard

1) Did you by any chance post it in this subforum because the numbers in the table are integers? (you just mentioned they were "numbers")

2) Good question grobber!

The numbers are real numbers!

3) You can have squares formed by many small squares For example, if you label the small squares in the obvious way, the cells for

a square of side The sum of the four numbers in these four cells has absolute value

4) so you can divide up each square?

Then I still don't get why it isn't infinite, like replace 10^9 by whatever in the below example

(where we split each square in 9 squares so all border squares fullfil the

requirement):

Perhaps can you reformulate it and/or make a drawing, cuz my english is too bad

to understand this I'm afraid

5) No, man, you cannot divide each square You have an infinite chess board

(without any coloring, though ) This board is made up of small squares of

fixed dimension However, from these small squares you can form larger squares,

made up of a certain number of small squares Is it that hard?

6) For example:

are small squares (you can't find smaller squares on your board)

However, the array formed by is another (larger) square The problem

7) then what does he mean by: 'an infinite field' and 'squares along the grid lines'?

If the fields is infinite, and you may just group together to larger squares, where are the grid lines then???

Gosh I'll better just quit this problem, i don't get the heck of it

8) Dude, think about the integer lattice That's naturally divided in small squares of side and it has grid lines of equations of the types and Let

be the number written in the cell numbered The problem says that (I'm just giving some examples, I can't write all of the relations because there are infinitely many ) It also says that

It also says that

, and so on (for all integers )

How on Earth can it still be unclear?

9) Cyclic equations

Turkey, 97

Prove that for each prime , there exists a positive integer n and integers

not divisible by p which satisfy

Hint

Lemma: for a prime there is s.t is a quadratic residue Proof: Take s.t (there is such a residue because )

Then satisfies the condition

10) IMO 1988/6, One of the hardest number theory problems

If are integers, then is a perfect square

Well I have seen many solutions of this problem Most assume to the contrary that q is not a perfect square Then I see nothing which follows uses this assumption (In fact I think the 'contradiction' holds even if we assume q to be a perfect square.) Then they choose a and b such that a+b is minimal Then they proceed further to prove that and In my opinion the proof goes exactly the same way had we assumed that

is a perfect cube, fifth power, etc

Will we not get the same 'contradiction' if we assumed to be a perfect square? Can anyone explain this to me?

Disscard

1) This is part of the solution I found in one of Titu's books

Let q be the number as stated in the problem We want to show that among all

pairs of nonnegative integers with the property and , the one with minimal has

But why? Can it be that some other and with not minimal have the property that is not a perfect square?

If you know the solution (or rather, understand the solution), can you please explain everything in detail?

2) Your problem is WRONG, because if then can not be a square, because is a negative number !!!!

3) The problem is not wrong It can be proved that if ab+1 were negative, the

resulting fraction can NEVER be an integer

By the way how did you type those mathematical equations?

4) There is a solution to this problem in Arthur Engel's book, involving hyperbolas

As has been said, the task is to show that if a solution exists for , then one exists for the same with , in which case must equal One shows that if a lattice point exists on the upper branch of the hyperbola , then a corresponding point exists directly below it Then we flip it across the axis

of symmetry and repeat until

Arthur Engel also rates it in his book (Problem Solving Strategies) as the hardest IMO problem until 1995 But I think this should probably be in the problem-solving section

5) Hmm Are you kidding? This is a very standart problem for a descent method And, of course, it can't be the hardest IMO problem (maybe easiest )

The same thing I can say about IMO95, pr6 (did Engle mean this problem?) 6) I still don't understand your solution Can you please explain in detail?

If is an integer, then is a perfect square

Can anyone teach me how to use Latex?

Engel meant IMO 1996/5, a geometry problem

Thank you

But I saw in some other sites that teach us to use some sort of \begin document blablabla What's the difference

And why is it that your line remains unchanged even when you use $?

7) [quote="Lek Huo Tang"]I still don't understand your solution Can you please explain in detail?

If $\frac{a^2+b^2}{1+ab} = q$is an integer, then $q$ is a perfect square

Can anyone teach me how to use Latex?

Engel meant IMO 1996/5, a geometry problem.[/quote]

You just need to use $ $ when you write formulas

11) Rational number

Find all numbers positive integer such that:

a) is rational number

b) is rational number

I haven't got any solution

Discard

1) Quite a few of these have been posted and solved around here Take the second one, for instance Let It's an algebraic integer, because it satisfies

This means that is also an algebraic integer, and so

is Since it's also rational, it means that it's integral, and it's easy from here: being in it can only be , and we solve it for these values 2) Well, I solved this problem but i can0t remember the method, but grobber failed, because In fact I think the unique solutions was Good luck looking for a nice solution for this nice problem

Now if you have carefully read, and it is clear that you did not, you would have seen that

grobber did not say that is the set of solutions for , but the values that

takes, and now we can easily solve each of the 5 equations

3) giving all the solutions, including the one you specified above

This is the Olympiad section of the site If needed more details on easy problems

such as this one, one can use the High-School forums

PS Next time I will not waste anymore time explaining posts of other users in such detail, as for kids, but I will rather directly delete the additional balast

12) Old Equation

Friend

Prove that the only solution in rational numbers of the equation:

is

Discard

1) this is a pretty standard method for this sort of problem (fermat's method of infinite descent)

must be divisible by 3, and from there we immediately get that

must be divisible by 27, which is to say that this then implies , which then implies that and so so if

is a solution, so is , and by induction

is a solution for all natural k and this is obviously impossible

2) A very beautiful problem, proposed for IMO by Titu Andreescu is

3) pleuristique: the problem asks for rational solutions

My guess is that the solution for this problem (and harazi's problem) involve the factorization of

4) Are you sure? If we multiply by the product of the denominators, we aren't

necessarily multiplying by a perfect cube

I solved it using the factorization

The expression factors into