Prove that there are no rational numbers x,y,z with x+y+z=0 and x2+y2+z2=100.
Problem
Source: Bundeswettbewerb Mathematik 2020, Round 2 - Problem 2
Tags: number theory, number theory unsolved, rational number, algebra, system of equations
17.11.2020 19:48
It same as proving, that there are not integer numbers a,b,c,d with a+b+c=0 and a2+b2+c2=d2 with d≠0 d2=a2+b2+c2=2(a2+b2+ab) But deg2(2(a2+b2+ab)) is odd, and deg2(d2) is even
17.11.2020 19:51
RagvaloD wrote: It same as proving, that there are not integer numbers a,b,c,d with a+b+c=0 and a2+b2+c2=d2 I don't think so. Your system has a solution (a,b,c,d)=(0,0,0,0) whereas the original problem has not.
17.11.2020 20:14
Yes, also d≠0 is needed
17.11.2020 21:33
let w be the least integer such that wx,wy,wz∈Z. We'll say a,b,c=wx,wy,wz respectively. We will also say that x,y,z≠0 It's then clear to see that gcd. We then have: a^2+b^2+c^2=100w^2a^2+b^2=100w^2-c^2a^2+b^2=(10w-c)(10w+c)Since c and w are integers, RHS cannot be congruent to 1 mod 4. Furthermore, since LHS is a sum of two squares, it cannot equal 3 mod 4. This means that they both must be even, implying that c is even. Since the problem is symmetric, the same argument can be applied to a and b. This contradicts \gcd(a,b,c)=1, therefore at least one of x y or z is zero. WLOG we say that z=0 We then have x^2+y^2=100x+y=0Solving this, it's clear to see that x and y are irrational, concluding the proof.
13.12.2020 22:36
Let's introduce the elementary symmetric functions of three variables which we denote by \sigma_{i}.. Then the given situation is interpreted as follows: We've \sigma_{1}=0 and \sigma_{1}^{2}-\sigma_{2}=50 \Leftrightarrow \sigma_{2}=-50 Therefore it follows that x,y,z are zeros cubic t^{3}-50t+k=0 where we've assumed that \sigma_{3}=xyz=k \in \mathbb{Q}(suppose) (Note: If xyz is non rational , we're done as then the cubic will have no real zeros) If r is a rational zero of the family of cubics of the above form, then by a fundamental and deep result it must be an integer, then we get r(50-r^2)=k.. From this we've the following cases: Case 1 (r,k)=(1,49) We can verify that the resulting cubic t^3 -50t +49=(t-1)(t^2 +t -49) has one real and two complex numbers as zeros Case 2 (r,k)=(7,7) We can similarly verify that the cubic t^2 -50t +7=(t-7)(t^2 +7t -1) has one real and two complex numbers as zeros.. Hence, there don't exist rational numbers satisfying the given conditions.. QED
16.02.2021 12:33
Is there a way to generalize this for 4 numbers a,b,c,d or even more?
16.02.2021 15:34
What do you want to replace the 100 with? If you have 4 integers a,b,c,d with a+b+c+d=0, then a^2+b^2+c^2+d^2=s reduces to (a+b)^2+(a+c)^2+(b+c)^2=s, which has or has no solutions according to the three squares theorem, assuming s is even. EDIT: But it is not impossible to generalize, see https://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=249323&start=0&lps=1822053#v1822053