Determine whether there exists a polynomial f(x1,x2) with two variables, with integer coefficients, and two points A=(a1,a2) and B=(b1,b2) in the plane, satisfying the following conditions: (i) A is an integer point (i.e a1 and a2 are integers); (ii) |a1−b1|+|a2−b2|=2010; (iii) f(n1,n2)>f(a1,a2) for all integer points (n1,n2) in the plane other than A; (iv) f(x1,x2)>f(b1,b2) for all integer points (x1,x2) in the plane other than B. Massimo Gobbino, Italy
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Tags: algebra, polynomial, function, ceiling function, inequalities, algebra proposed
14.05.2010 09:22
Can anybody please provide solutions or ideas?
15.05.2010 10:07
Here's some ideas on how you might go about finding a solution. Let's say that for example we're trying to take B=(1005.5,1004.5), A=(0,0). Our goal will be to construct a polynomial such that (1) f(x,y)≥0 everywhere. (2). f(1005.5,1004.5)=0. Nowhere else is f equal to 0. (3). f(0,0), while nonzero, is still the smallest value taken on by f on an integer. We can satisfy (1) by making sure we right f(x,y) as a product of polynomials which are either squares or sum of squares. So we just need to choose the factors right. To satisfy (2): Find a polynomial which is 0 only at (1005.5,1004.5), and nowhere else. As long as this is one of your factors, and none of your factors have real roots, you should be good. To satisfy (3): You want to look a polynomial which does not itself have real roots, but is HUGE at every integer except for (0,0) (HUGE in this case means that f(0,0) is so much smaller than the other values that it dwarfs any contribution from the factor you found in the previous step). Once you find the two polynomials meeting the criteria above, take their product and you'll have one that meets all the original criteria.
06.07.2010 12:54
here is may solution let A(0,0)B(14,80394) and f(x1,x2)=(x1−14)2+(8039x1−x2)2+34 and we can easily prove that this function is suitable
12.10.2014 16:08
A slight correction to the above post: You can use the same idea, but a better polynomial to use is (4x1−1)2+(8039x1−x2)2. A=(0,0) and B=(14,80394) still. Firstly, it's obvious that this is an integer polynomial. Secondly, f attains a minimum of 0 at only the point B (you show this by setting the things inside parentheses to equal 0). Since f is an integer polynomial, for any integer point X we have f(X)≥1 (since I showed it doesn't equal 0). Note that f(A)=1. So it suffices to show that f attains the value 1 only at A. We firstly need for (4x1−1)2≤1⟹x1=0 (remember that everything is an integer point). So (4x1−1)2=1, so we need to (8039x1−x2)2=0⟹x2=0. So the only integer point that achieves the value 1 is As, so we're done.
01.02.2015 21:34
20.02.2015 06:16
How about f(x,y)=((x2+y2−(2009.92+0.12))(x2+y2−ϵ))2 where ϵ is insanely small. Of course, A=0,B=(2009.9,0.1). Let r=0.12+2009.92, not an integer. First, notice that f(2009.9,0.1)=0 and so it's the smallest possible value f attains. Now, we want f(0,0)=((ϵ)r)2<(X−ϵ)2(X−r)2 for any X positive integer. So we want |ϵX−ϵ|<|X−rr|. Notice that the the RHS is bounded below, while the LHS is bounded above, so we can make the LHS as small as we want with a correct ϵ.
14.02.2016 18:31
WLOG A is the origin. We want a polynomial f∈Z[x1,x2] which attains its minimum over Z2 at A, such that its value at B is smaller. The conditions are very weak, so yes, this is a construction problem. Say we just want a minimum at A. Then we can just take the distance from A squared, x2+y2, that's minimal at A. The joke is that B is not necessarily an integer point. For example, the polynomial t2−t is negative in ]0;1[, but non-negative elsewhere. Seriously abusing this, we choose B=(2009.5,0.5) and set f(x,y)=(x2+y2)+googolplex⋅(y2−y). Then f(x,y)=0 for (x,y)=A, but f(x,y)>0 for any other integer point. However, f(B) is absurdly negative because 0.52−0.5=−14.
27.09.2016 23:31
Define the polynomial g(X,Y) as follows: g(X,Y)=(X2+Y2)⋅(2⋅(X2+Y2−20102+1)2−1)and let us set f(X,Y)=g(X−a1,Y−a2). Let c,d be positive real numbers such that c+d=2010 and c2+d2=20102−1. It is clear that such numbers c,d exist and neither is an integer. Let b1−a1=c and b2−a2=d. Clearly, g has integer coefficients and for all non zero lattice points (a,b) we have a2+b2>0 and since 20102−1=2009⋅2011 and 2011 is a prime number of the form 4k+3 with exponent 1 in this expression, we see that a2+b2≠20102−1. Thus, g(a,b)>g(0,0)=0 and as c2+d2=20102−1 we get that g(c,d)<0 and thus, f satisfies all the desired conditions.
14.02.2018 05:21
Let g(t)=t(3t−(3k+1))(3t−(3k+2)), where k is a positive integer. Note that g is minimized on nonnegative integer inputs at t=0 and minimized on nonnegative reals at some t=a∈(3k+1,3k+2). In particular g(t)>0 for all t∈N. Let f(x,y)=g(x2+y2), and set k to be some positive integer such that the circle x2+y2=a2 intersects the diamond |x|+|y|=2010 at some point (b1,b2). Then, we take (a1,a2)=(0,0), and note that all conditions are satisfied.
14.01.2019 23:48
Does this work? f(x,y)=((2x−2011)2+(2y−2009)2)(C(x2+y2)+1) A=(0,0) B=(2011/2,2009/2) for C sufficiently large