Advanced book on Mathematics Olympiad

Geometry and Trigonometry 671 Using the identity sina−cosa= √ 2 sin(a− 45 ◦)in the numerators, we transform this further into √ ...

672 Geometry and Trigonometry Because sin^8120 π=sin( 4 π+ 20 π)=sin 20 π, this is equal to 161. (T. Andreescu) 698.(a) We obser ...

Number Theory 699.Becausean− 1 ≡n− 1 (modk), the first positive integer greater thanan− 1 that is congruent tonmodulokmust bean− ...

674 Number Theory 701.Fromf( 1 )+ 2 f(f( 1 ))=8 we deduce thatf( 1 )is an even number between 1 and 6, that is,f( 1 )= 2 ,4, or ...

Number Theory 675 −an≤−b^2 n≤−an− 1 , an≤b^2 n+ 1 ≤an+ 1. Adding these two inequalities, we obtain 0 ≤bn^2 + 1 −bn^2 ≤bn+bn− 1 , ...

676 Number Theory and ifx= 2 k+1 withk>3, then x^3 =( 2 k+ 1 )^3 =( 2 k− 1 )^3 +(k+ 4 )^3 +( 4 −k)^3 +(− 5 )^3 +(− 1 )^3. In ...

Number Theory 677 So 3 dividesz^2 +t^2. Since the residue of a square modulo 3 is either 0 or 1, this can happen only if bothzan ...

678 Number Theory 710.Assume that the property does not hold, and fixa. Only finitely many numbers of the formf(a+k)can be less ...

Number Theory 679 Let us verify first that 0<r. Assume the contrary. Sincebr=a^2 −ka+1, we must havea^2 −ka+ 1 ≤0. The equali ...

680 Number Theory x 0 =m^2 +n^2 ,x 0 +y 0 − 2 z 0 =p^2 +q^2 ,z 0 −x 0 =mp+nq. In that case, y 0 =p^2 +q^2 + 2 z 0 −x 0 =p^2 +q^2 ...

Number Theory 681 and, according to Hermite’s identity, Sn−Sn− 1 = ⌊ n x n ⌋ =x. BecauseS 1 =x, it follows thatSn=nxfor al ...

682 Number Theory (Korean Mathematical Olympiad, 1997) 718.The functionf:[ 1 ,n(n 2 +^1 )]→R, f(x)= − 1 + √ 1 + 8 x 2 , is, in f ...

Number Theory 683 As before, we deduce that 2 m+^1 b= 2 m+^1 b+ba. Again this is an impossibility. Hence the only possibilit ...

684 Number Theory or rn≤ r 1 1 + r 2 2 + r 3 3 +···+ rn n . Truncate the sum on the right to the(q− 1 )st term. Sincepandqare co ...

Number Theory 685 This proves the claim and completes the solution. (34th International Mathematical Olympiad, 1993) 723.Suppose ...

686 Number Theory Remark.The casek=1 was given at the 20th International Mathematical Olympiad, 1978; the idea of the solution w ...

Number Theory 687 Because 2a−1 and 2a+1 are coprime, and so are 2b−1 and 2b+1, this is further equal to gcd( 2 a− 1 , 2 b− 1 )·g ...

688 Number Theory for some integersa, b, c. Identifying coefficients, we must have a+b=− 4 , c+ab+ 1 = 6 , b+ac+ 1 =− 4 , a+c= 1 ...

Number Theory 689 f( 2 n+ 1 )= 3 f (n)+ 1 , f( 2 n)= 3 f (n). At this moment it is not hard to guess the explicit formula forf; ...

690 Number Theory 735.The numbersxandyhave the same prime factors, x= ∏k i= 1 pαii,y= ∏k i= 1 pβii. The equality from the statem ...