Advanced book on Mathematics Olympiad

390 Algebra P (m)= (− 1 )m+^1 m(m− 1 )···(m−n) (n+ 1 )!(m+ 1 ) + m m+ 1 . (D. Andrica, published in T. Andreescu, D. Andrica, 36 ...

Algebra 391 whereQ(x)=b(x 1 −x)(x 2 −x)···(xn−x). This completes the proof. (Romanian Mathematical Olympiad, 1979, proposed by M ...

392 Algebra a+b+c=r, ab+bc+ca=r, abc= 1. We deduce thata, b, care the solutions of the polynomial equationt^3 −rt^2 +rt− 1 =0. T ...

Algebra 393 3 =x 12 +x 22 +···+xn^2 ≥nn √ x^21 x 22 ···xn^2 =n. Therefore,n≤3. Eliminating case by case, we find among linear po ...

394 Algebra tan(α+β+γ)= tanα+tanβ+tanγ−tanαtanβtanγ 1 −(tanαtanβ+tanβtanγ+tanαtanγ) , implies tan(α+β+γ)= u+v+w−uvw 1 −(uv+vw+uv ...

Algebra 395 holds true forα= 2 nπ+ 1 , 2 n^2 π+ 1 ,..., 2 nnπ+ 1. Hence the equation ( 2 n+ 1 1 ) xn− ( 2 n+ 1 3 ) xn−^1 +···+(− ...

396 Algebra ≤ 1 27 (x+α+β+γ)^3 = 1 27 (x−a)^3 , so thatP(x)≥− 271 (x−a)^3. Equality holds exactly whenα−x=β−x=γ−xin the first in ...

Algebra 397 13 n+^1 ≡ 13 (mod 8)sincenis even, a contradiction. We conclude thata= 13 ,n= 2 is the unique solution. (62nd W.L. P ...

398 Algebra 168.Assume without loss of generality that deg(P (z))=n≥deg(Q(z)). Consider the polynomialR(z)=(P (z)−Q(z))P′(z). Cl ...

Algebra 399 in the ring of polynomials with integer coefficients. Since the two polynomials are monic and have integer coefficie ...

400 Algebra IfP′(x^1 + 2 x^2 )=0, then this identity gives 0 = 1 x 1 +x 2 2 −x^3 + 1 x 1 +x 2 2 −x^4 +···+ 1 x 1 +x 2 2 −xn < ...

Algebra 401 An exercise in the section on induction shows that for any positive integerk,|sinkt|≤ k|sint|. Then rn= sint sin(n− ...

402 Algebra z+z 1 z−z 1 + z+z 2 z−z 2 +···+ z+zn z−zn have negative real part, while ifzhas absolute value greater than 1, all t ...

Algebra 403 180.The hypothesis of the problem concerns the coefficientsamanda 0 , and the conclusion is about a zero of the poly ...

404 Algebra Returning to the problem, we see that ify 0 andy 1 are zeros ofP′(x)−kP (x)with y 0 <y 1 , then they are separate ...

Algebra 405 (Q+R)(ai)=0,i = 1 , 2 ,...,n. Since theai’s are all distinct and the degree of Q(x)+R(x)is at mostn−1, it follows th ...

406 Algebra a contradiction. We conclude thatP(x)is irreducible. 188.Assume the contrary, and let (x^2 + 1 )n+p=Q(x)R(x), withQ( ...

Algebra 407 190.Letnbe the degree ofP(x). Suppose that we can find polynomials with integer coefficientsR 1 (x)andR 2 (x)of degr ...

408 Algebra Now let us assume thatP(x)is a polynomial for whichM=^18. Thenb=−1, d=^18. Writing the double inequality−^18 ≤P(x)≤^ ...

Algebra 409 2 (n−^1 )(n−^2 )/^2 ∏ 1 ≤i<j≤n (cosjα−cosiα) = 0. 196.Because the five numbers lie in the interval[− 2 , 2 ], w ...