Problem 1. LetABC and AM N be two similar triangles with the same ori-entation, such thatAB = AC, AM = AN , and having disjoint interiors. Let O be the circumcenter of the triangleM AB. Prove that the points O, C, N , A are concyclic if and only if the triangleABC is equilateral.
Solution. Letα = ∠BAC = ∠M AN . We consider the rotation of center A and angle α; from the hypothesis we infer that B is mapped onto C, and M is mapped ontoN . This means that the triangle BAM is transformed into the triangleCAN , and thus O is mapped onto O0, the circumcenter of the triangle CAN . Moreover, ∠OAO0 = α and OA = O0A.
The condition thatO, C, N , A lie on the same circle is equivalent to O0O = O0A (as already O0is the circumcenter of the triangleCAN ). But then the triangle O0AO is equilateral, therefore α = 60◦, and the trianglesABC and AM N are also equilateral. The above reasoning works both ways, so the problem is solved.
Problem 2. Letp > 5 be a prime number. Find the number of irreducible polynomials inZ[X], of the form
xp+ pxk+ pxl+ 1, k > l, k, l∈ {1, 2, . . . , p − 1} .
Solution. Letfk,l(x) = xp+ pxk+ pxl+ 1, k > l, k, l ∈ {1, . . . , p − 1}.
If the numbersk and l have different parities, then fk,l(−1) = 0. For fk,l(x) to be irreducible inZ[X] it is required that the numbers k and l have the same parity;
then
fk,l(x− 1) = xp+ pxg(x) + p((−1)k+ (−1)l) = xp+ pxg(x)± 2p,
whereg(x)∈ Z[X].
By Eisenstein’s criterion, fk,l(x− 1) is irreducible in Z[X], thus fk,l(x) is irreducible inZ[X].
Therefore the number of polynomialsfk,l(x), irreducible in Z[X], is equal to the number of pairs(k, l), in which k, l are distinct numbers, of the same parity, in the set{1, 2, . . . , p − 1}. The number of such pairs is
2
p−1
2
2
= (p− 1)(p − 3)
4 .
Remarks. The idea of making the transformationx7→ x − 1 is suggested by the method used by Gauss to prove the irreducibility of thep-th order cyclotomic polynomial
φp(x) = xp−1+ xp−2+· · · + x + 1.
Problem 3. Leta, b be positive integers such that for any positive integer n we havean+ n| bn+ n. Prove that a = b.
Solution. Assume thatb6= a. Taking n = 1 shows that a + 1 divides b + 1, henceb > a. Let p > b be a prime and let n be a positive integer such that
n≡ 1 (mod (p − 1)) and n ≡ −a (mod p).
Such ann exists by the Chinese Remainder theorem (without the Chinese Remain-der theorem, one could notice thatn = (a + 1)(p− 1) + 1 has this property).
By Fermat’s little theorem,an = a(ap−1· · · ap−1) ≡ a (mod p), and there-forean+ n≡ 0 (mod p). So p divides the number an+ n, hence also bn+ n.
However, by Fermat’s little theorem again, we have analogouslybn+ n≡ b − a (mod p). We are therefore led to the conclusion p| b−a, which is a contradiction.
Remarks. The first thing coming to mind is to show thata and b share the same prime divisors. This is easily established by using Fermat’s little theorem or Wilson’s theorem. However, we know of no solution which uses this fact in any meaningful way.
For the conclusion to remain true, it is not sufficient thatan+ n| bn+ n holds for infinitely manyn. Indeed, take a = 1 and any b > 1. The given divisibility
relation holds for all positive integersn of the form p− 1, where p > b is a prime, (b) Prove that forn > 2 the bound above is the best possible.
Solution. (a) We may suppose thata1 > 0, otherwise if a1 = 0, k = 1 is a
Becauses1 = a1 > 0 it follows that there exists (a smallest) k such that sk< 0. Then in contradiction with the previous relation.
(b) We will treat separately the cases whenn is odd and n is even.
CASEI.n odd (n > 1). We take Then we obtain the sequence
{si}i=1,n= 3
CASEII.n even (n > 2). We take
Then we obtain the sequence
{si}i=1,n= disproving the possibility of lowering the bound.
Problem 5. Let{an}n>1be a sequence given bya1= 1, a2= 4, and for all integersn > 1
an=p
an−1an+1+ 1.
(a) Prove that all the terms of the sequence are positive integers.
(b) Prove that the number 2anan+1 + 1 is a perfect square for all integers n > 1.
Solution. (a) We rewrite the recurrence relation as an+1= a2n− 1
an−1
and we want to prove using induction thatak∈ N for all k 6 n implies an+1∈ N.
For this we require the following, stronger, statement: ak ∈ N, ∀k 6 n, and in plusgcd(ak, ak−1) = 1 for all k 6 n. The initial two steps n = 2 and n = 3 are easily calculated, so we suppose thatn > 4.
The relationan=a
(b) Taking small values forn we notice that
2anan+1+ 1 = (an+1− an)2,
and we will prove this relation using induction. The first step,n = 1, is trivial. Let n > 2. The relation implies
2anan+1= a2n+1− 2anan+1+ a2n− 1 = an+1(an+1− 2an) + an+1an−1, therefore, by dividing both sides above withan+1 > 0, we obtain the equivalent relation
4an= an+1+ an−1 ⇒ an+1 = 4an− an−1. We only have to prove thatan+2 = 4an+1 − an. Butan+2 = a
2 n+1−1
an so we require
4an+1an− a2n = a2n+1− 1 ⇔ 2anan+1+ 1 = (an+1− an)2, which is our induction hypothesis, and we are done.
ALTERNATIVESOLUTION. (a) First notice that the sequence may be extended to the left withx0= 0 (this helps with having the recurrence relation also available forn = 1). Now, writing two consecutive (squared) recurrence relations yields, forn > 1, x2n = xn+1xn−1+ 1 and x2n+1 = xn+2xn+ 1, so by subtracting, xn+1(xn+1+ xn−1) = xn(xn+2+ xn), that is,
xn+2+ xn
xn+1
= xn+1+ xn−1 xn
,
thus having constant value(x2+x0)/x1= 4, whence xn+1= 4xn−xn−1(clearly xn6= 0 for n > 1).
Therefore, once the first two terms are given as integers, so will all following terms be.
(b) We have, forn > 1, 0 = xn+1(xn+1−4xn+ xn−1) = x2n+1−4xn+1xn+ xn+1xn−1= x2n+1−4xn+1xn+ x2n−1 = (xn+1−xn)2−(2xnxn+1+ 1), hence 2xnxn+1+ 1 = (xn+1− xn)2and therefore a perfect square.
Remarks. The first solution entails a longer and more ardous process, which fails to provide the linear recurrence that turns to be instrumental in proving (b).
One may (similarly) easily obtain that 3x2n + 1 = (xn+1 − 2xn)2, there-fore a perfect square, thus falling over a solution family for the Pell equation y2− 3x2= 1.
This type of sequences and the way to attack them is quite well-known, see [A. Engel], [A. Negut¸], [V. Vornicu].
Problem 6. LetABC be a triangle with ∠ABC = 30◦. Consider the closed discs of radiusAC/3 centered at A, B and C. Does there exist an equilateral triangle whose three vertices lie one each in each of the three discs?
Solution. We will start with the following
LEMMA. Given two pointsA1, A2and two closed discs centered atA1, A2
of radiir1, r2respectively, the locus of the third vertex of an equilateral triangle with the two other vertices lying in each of the two discs and located in one of the halfplanes determined by the lineA1A2 is the closed disc of radiusr1+ r2
centered atA3which forms withA1andA2an equilateral triangle.
To be self-contained, we present a proof using complex numbers. Let lower-case letters represent the affixes of capital-letter points. Thena3 = a1+ ω(a2− a1) = ¯ωa1+ ωa2, whereω = cos 60◦+ i sin 60◦is the primitive6-th root of 1, hence|ω| = 1, ¯ω = 1− ω and ω ¯ω = 1 (if in the other halfplane, we may use the symmetrical relationa3= ωa1+ ¯ωa2).
Take nowa01= a1+ α1,a02= a2+ α2; thena03= ¯ωa01+ ωa02= a3+ (¯ωα1+ ωα2) = a3+ α3. For|α1| 6 r1,|α2| 6 r2we get|α3| = |¯ωα1+ ωα2| 6 r1+ r2.
Conversely, for anyα3 with|α3| 6 r1+ r2one can takeα1 = rr1
1+r2ωα3, α2= rr2
1+r2ωα¯ 3and get|α1| 6 r1,|α2| 6 r2andωα¯ 1+ ωα2= α3.
C
C’ B C’’
B’’
B’
A’
30
A’’
A
Returning to the original problem and denoting byC0,A0,B0 the points that form an equilateral triangle with(A, B), (B, C), (C, A) respectively, in the same halfplane withC, A, B respectively, one easily establishes that AC = AA0 = CC0 = BB0. This is becauseA lies on the perpendicular bisector of CA0,C lies on the perpendicular bisector ofAC0, while ∠ABC = 30◦and ∠AB0C = 60◦ implies that B0 is the circumcenter of the triangle ABC, hence B0B = B0C = AC.
By the lemma, the pointsA00 onAA0 such thatAA00 = AA0/3, C00onCC0 such thatCC00 = CC0/3 and B00onBB0such thatBB00 = BB0/3 are the only points available as vertices for an equilateral triangle like that asked for, as the discs of radii 23AC centered at A0,C0,B0 are in fact tangent to the discs of radii
1
3AC centered at A, C, B respectively.
We have proved that there exists such a triangle, and in fact that the triangle is unique.
Problem 7. Determine the pairs of positive integers(m, n) for which there exists a setA such that for x, y positive integers, if|x − y| = m, then at least one of the numbersx, y belongs to the set A, while if|x − y| = n, then at least one of the numbersx, y does not belong to the set.
Solution. Fork positive integer, we will denote by ν(k) the exponent of 2 in the decomposition in prime factors ofk. We shall prove that the pairs (m, n) that fulfill the hypothesis are the ones for whichν(m) = ν(n).
Let us suppose that a setA with the properties in the hypothesis exists; then for a∈ A we have a+n /∈ A thus (a+n)+m ∈ A which means ((a+n)+m)−n /∈ A, thereforea + m /∈ A.
Analogously, forb /∈ A we have b + m ∈ A thus (b + m) + n /∈ A which means that((b + m) + n)− m ∈ A, therefore b + n ∈ A.
Therefore, forx, y with|x − y| ∈ {m, n}, one of them belongs to A and the other one does not, so the problem’s statement is symmetric inm, n.
Through a simple induction we obtain thata+km and a+kn both belong to the same set (A or its complementary) as a for k even, and to different sets for k odd.
Let us suppose now thatν(m) 6= ν(n). Without loss of generality we may
suppose thatν(m) > ν(n). Then m = 2ν(m)· m0,n = 2ν(n)· n0, withm0, n0
It is easy to verify that this set fulfills the conditions in the statement.
Problem 8. Letxi,1 6 i 6 n be real numbers. Prove that
Solution. The inequality above is equivalent with
X
We may suppose that (if necessary by reindexing the variablesxi) xi>0, 1 6 i 6 k, xj < 0, k + 1 6 j 6 n. otherwise we work with−xiinstead. We have
X
Therefore,
As an aside, it is illuminating that2kp + 2(n− k)m > n(p + m), equivalent to(2k− n)(p − m) > 0 is not necessarily true (e.g., for 2k > n, p < m take
ALTERNATIVESOLUTION. From the obvious relation, fora, b∈ R,
|a| + |b| − |a + b| =
6
which is equivalent to the stated inequality.
Problem 9. The circle of center I is inscribed in the convex quadrilateral ABCD. Let M and N be points on the segments AI and CI respectively, such that ∠M BN =12∠ABC. Prove that ∠M DN = 12∠ADC.
Solution. Denote by ∠A, ∠B, ∠C, ∠D the angles of ABCD. Since BI is the bisector of ∠ABC and ∠M BN = 12∠B, we may put
Consider trianglesAM B and M BI. From the sine theorem it follows that AM
sin α1 = BM
sin A2 , M I
sin α2 = BM sin ∠BIM.
Thus
Similarly, from considering trianglesAM D and M ID we have sin β1
Similarly considering trianglesIBN and N BC, and then triangles IDN and N DC, we get
Now consider trianglesABI and DIC:
∠BIA = 180◦−∠A
It follows thatsin ∠BIA = sin ∠DIC and similarly sin ∠AID = sin ∠BIC.
Multiplying(∗) and (∗∗) we obtain sin β1
Problem 10. LetA be a point exterior to a circleC. Two lines through A meet the circleC at points B and C, respectively at D and E (with D between A and E). The parallel through D to BC meets the second time the circleC at F . The lineAF meetsC again at G, and the lines BC and EG meet at M. Prove that
1
AM = 1
AB + 1 AC.
Solution. Since lines DF and AC are parallel, it follows that the angle
∠DF A = ∠CAF . On the other hand, ∠DF A = ∠DEG because both an-gles subtend the arcDG. Thus ∠CAF = ∠DEG, whence triangles AM G and EM A are similar, which implies AMM G =EMAM, that isAM2= M G· ME.
D
A M B C
G E
F
By the power of a point theorem,M G· ME = MB · MC, whence AM2= M B· MC = (AB − AM)(AC − AM) = AB · AC − AM(AB + AC) + AM2, which yieldsAM (AB + AC) = AB· AC, equivalent to the required relation.
Remarks. The condition “D between A and E” is required, not so much be-causeAF may be tangent toC (in which case G ≡ F , but the result holds), but because the parallel throughD to BC may be tangent toC (in which case D ≡ F , whenceE≡ G, and the line EG does not exist). Otherwise, except this degenerate case, the result holds also whenE between A and D.
Problem 11. Letγ be the incircle of the triangle A0A1A2. In what follows, indices are reduced modulo3. For each i∈ {0, 1, 2}, let γibe the circle through Ai+1 andAi+2, and tangent toγ ; let Tibe the tangency point ofγi andγ ; and finally, letPibe the point where the common tangent atTitoγiandγ meets the lineAi+1Ai+2. Prove that
(a) the pointsP0,P1andP2are collinear;
(b) the linesA0T0,A1T1andA2T2are concurrent.
Solution. (a) Consider the power ofPirelative to the circlesA0A1A2,γiand γ: the first equals PiAi+1· PiAi+2; the second equalsPiAi+1· PiAi+2 = PiTi2; and the third equalsPiTi2. Consequently,Pilies on the radical axis of the circles A0A1A2andγ and we are done.
A X(ABCD) for the cross-ratio of the rays XA, XB, XC, XD.
By Brianchon’s theorem,Ai+1Vi∩ SiTi∩ Ai+2Ui= ωi. Now,
where (1) holds by Menelaus’ theorem using (a), and (2) holds becauseSiAi+1= Si+2Ai+1as tangents toγ from Ai+1. The conclusion follows by the converse to Ceva’s theorem.
ALTERNATIVE SOLUTION TO(b). Denote by4T the triangle made by the three common tangents; its sides are the tangents through the vertices of the
trian-gleT0T1T2to its circumscribed circle, whence by Lemoine’s theorem, their inter-sections with the sides of triangleT0T1T2are collinear (Lemoine’s line). Now, De-sargues’ theorem shows that4T and triangle T0T1T2are perspective. By (a), De-sargues’ theorem similarly shows that4T and triangle A0A1A2are perspective.
But the relation (for triangles) of being perspective is transitive, hence triangles A0A1A2andT0T1T2are perspective and therefore the linesAiTiare concurrent.
It is obvious, from both solutions, that this problem calls for projective methods to be used.
Solution. Letx = ab + bc + ca. From the well-known inequalities (a + b + c)2>3(ab + bc + ca) and (ab + bc + ca)2>3abc(a + b + c)
Equality occurs if and only ifa = b = c = 1.
ALTERNATIVE SOLUTION. Let us make the notation f (x) = 1
= P(1 − a)f(a). Because of the symmetry of the relation we may assume without loss of generalitya > b > c; then using 1− c = (a− 1) + (b − 1) and 1 − a = (c − 1) + (b − 1) we get
This shows thatf is decreasing. All that is left now is to use relation (∗) for b > 1 and relation (∗∗) for b 6 1 .
Remarks. Yet another solution would be to homogenize the equation and use
“brute force” along with AM-GM and Muirhead, as in one of the solutions of IMO 2005, Problem 3.
The real beautiful thing to say is that if instead of 3 variables we think of the inequality withn variables
this holds up ton = 10. For n > 11 it fails, e.g., for x1 = · · · = x10 = 0.6, x11 = 5, and xi = 1, for all i > 12. The proof for 4 6 n 6 10 involves mixed variables (Sturm-type) techniques and is probably worth a short article by itself.
Problem 13. Givenr, s∈ Q, determine all functions f : Q → Q such that f (x + f (y)) = f (x + r) + y + s
for allx, y∈ Q.
Solution. Denoteg(x) = f (x)− r − s. The functional equation becomes g(x + g(y)) = g(x + f (y)− r − s)
= f ((x + r− s) + f(y)) − r − s
= f (x− s) + (y − s) − r − s
= g(x− s) + y + s, and
g2(x + g(y)) = g(y + s + g(x− s)) = g(y) + x − s + s = x + g(y), henceg2= id, on elements of the form x + g(y). By fixing y = y0, the set
{x + g(y0) : x∈ Q} = Q, henceg2= id on all of Q, so g is one-to-one.
Then, by replacingy by g(y) (any y is in the image)
g(x + y) = g(x + g(g(y))) = g(x− s) + g(y) + s.
Obviously,g(y + x) = g(y− s) + g(x) + s, hence g(x) − g(x − s) = g(y) − g(y− s) = kga constant. This gives
g(x + y) = g(x) + g(y) + (s− kg).
Forx = y = 0: s− kg=−g(0), so g(x + y) = g(x) + g(y) − g(0). Consider all solutionsg having g(0) = z fixed; then for any two such solutions g1,g2, denoting
h = g1− g2we geth(x + y) = h(x) + h(y), so h(x) = λx (with λ = h(1)).
On the other hand,g0(x)≡ z is obviously a particular solution, hence the general solutiong is g(x) = h(x) + g0(x) = λx + z.
Checking this into the main relation forg, we get
g(x + g(y)) = g(x + λy + z) = λx + λ2y + λz + z and
g(x− s) + y + s = λx − λs + s + z + y + s = λx + y − λs + s + z, henceλ2= 1, λz =−λs + s = s(1 − λ). We have therefore two solutions:
λ = 1; z = 0 gives g(x) = x;
λ =−1; z = −2s gives g(x) = −x − 2s.
These lead to the following solutions forf :
f (x) = x + r + s and f (x) =−x + r − s.
Remarks. As it is, the method of solving an equation
ϕ(x + y + a) = ϕ(x) + ϕ(y) + a,
is refreshingly reminiscent of solving Cauchy, combined with the theory of ho-mogenizing plus a particular solution.
Problem 14. Find all positive integersm, n, p, q such that pmqn = (p + q)2+ 1.
Solution. Clearly we havep| q2+ 1 and q| p2+ 1. Now, if we assume p = q, it followsp| p2+ 1, so p = q = 1 which is not a solution.
We may therefore assume without loss of generalityp < q. But p = 1 leads to q = 2, and p = 1, q = 2 is again no solution, therefore 2 6 p. We have
pmqn= (p + q)2+ 1 < 4q26p2q2< pq36pmq3,
son < 3. The case n = 2 leads to pm< 4, so m = 1, p = 2 or p = 3. For p = 3 we get3q2= (3 + q)2+ 1, impossible, while for p = 2 we get 2q2= (2 + q)2+ 1, whenceq = 5, a solution.
The case n = 1 leads to pm < 4q. Now, from q | p2+ 1, if q = p2+ 1 thenp | q2+ 1 = p4+ 2p2+ 2 implies p = 2, whence q = 5, but 2m· 51 = (2+5)2+1 = 50 is impossible, so q 6 p2+ 1
2 , whencepm< 2(p2+1), therefore p = 2 and m 6 3, or m 6 2. But p = 2 leads to q 6 5
2, soq 6 2, impossible, hence trulym 6 2.
Form = 1 the equation writes
pq = (p + q)2+ 1, clearly impossible.
Form = 2 we get p2 < 4q which combined with q | p2+ 1 leads to the possibilitiesp2+ 1 = q, 2q, 3q, 4q. But the case p2+ 1 = q was dismissed in the above, whilep2+ 1 cannot be congruent with 0 modulo 3, nor 4, so the only case left isp2+ 1 = 2q. But then, from p| q2+ 1 we get 4p| p4+ 2p2+ 5, so p| 5, i.e.,p = 5, q = 13, which checks as a solution.
Therefore, the only solutions are
(m, n, p, q)∈ {(2, 1, 5, 2), (2, 1, 5, 13), (1, 2, 2, 5), (1, 2, 13, 5)}.
If anyone reaches the equationp2q = (p+q)2+1 (dealing with the preamble in any different manner), yet another way to solve this is to consider it as a polynomial inq whose discriminant ∆ = (2p−p2)2−4p2−4 = (p2−2p−2)2−8p−8 should be non-negative, which impliesp > 5, and should also be a perfect square. But the next lower perfect square is(p2− 2p − 3)2= (p2− 2p − 2)2− 2(p2− 2p − 2) + 1, thus we must have−2(p2− 2p − 2) + 1 > −8p − 8, that is 2p2− 12p − 13 6 0, which impliesp 6 6.
Finally, we only have to check it for two values,p = 5 and p = 6, and only the former provides a solution (see above).
This fruitful combination of divisibility constrains which lead to inequalities is also reminscent of the BMO 2005, Problem 2.
ALTERNATIVESOLUTION. The “proof from the Book” solution makes use of the following known
Lemma. The equationx2+ y2+ 1 = kxy has (infinitely many) solutions in positive integers if and only ifk = 3.
Indeed, assumek 6= 3, then a solution x0,y0cannot havex0 = y0; without loss of generality we may takex0< y0. One then also has
(kx0− y0)2+ x20+ 1 = k(kx0− y0)x0,
as it may be readily verified. Butkx0y0 = x20+ y20+ 1 > y02implieskx0> y0, whilekx0y0= x20+ y02+ 1 < 2y02implieskx0< 2y0, therefore0 < x0(kx0− y0) < x0y0.
Takex1= min{x0, kx0−y0}, y1= max{x0, kx0−y0}; one has 0 < x1y1<
x0y0, which by Fermat’s infinite descent method leads to a contradiction.
Conversely, fork = 3, we have the infinite family of solutions (1, 1), (1, 2), (2, 5), (5, 13), . . . , (xn, yn), . . ., with xn+1= yn,yn+1 = 3yn− xn.
For an alternative solution to the lemma, using Pell equation techniques, see [A. Gica, L. Panaitopol].
Back to the original problem, the stated equation may be written as p2+ q2+ 1 = (pm−1qn−1− 2)pq.
According to the lemma above, in order to have positive integer solutions, pm−1qn−1− 2 = 3, that is, pm−1qn−1= 5.
This quickly leads to the solutions(m, n, p, q) ∈ {(2, 1, 5, 2), (2, 1, 5, 13), (1, 2, 2, 5), (1, 2, 13, 5)}.
Remarks. While in English the expression “positive integern” means an in-tegern > 0, in Romanian the verbatim translation of the expression includes the casen = 0; this leads to a case which was not meant to be considered, but which is worth noticing to be also true: formn = 0 the equation is equivalent to xr= y2+1 which has no solutions, either by invoking Catalan’s theorem, or by direct proof in Gauss’ integer ringZ[i] using parity arguments.
Problem 15. Letn > 1 be an integer. A set S⊂ {0, 1, . . . , 4n − 1} is called sparse if for anyk∈ {0, 1, . . . , n − 1} the following two conditions are satisfied:
(1) the setS∩ {4k − 2, 4k − 1, 4k, 4k + 1, 4k + 2} has at most two elements;
(2) the setS∩ {4k + 1, 4k + 2, 4k + 3} has at most one element.
Prove that the set{0, 1, . . . , 4n − 1} has exactly 8 · 7n−1sparse subsets.
Solution. It is enough to have available a set of 7 elements (at the “end” of the set) in order to write some recurrence relations.
[4n− 3 4n − 2 4n − 1] 4n [4n + 1 4n + 2
| {z }
4n + 3].
Denote byTnthe total number of sparse sets, byAnthe number of sparse sets that contain one of the “last” two elements (4n− 1, 4n − 2) and by Bn the number of sparse sets that contain none of these two elements (no sparse set may contain both because of condition (2)).
Then
An+1= Tn(4n + 3 and no other element > 4n) + Tn(4n + 3 and 4n)
+ Tn(4n + 2 and no other element > 4n) + Bn(4n + 2 and 4n)
= 3Tn+ Bn, and
Bn+1 = Tn(no elements > 4n) + Tn(4n + 1, but not 4n) + Tn(4n, but not 4n + 1) + Bn(both 4n and 4n + 1)
= 3Tn+ Bn,
henceAn+1= Bn+1andTn+1 = An+1+ Bn+1 = 6Tn+ 2Bn.
Now it is enough to calculateA1andB1; clearly{2}, {0, 2}, {3}, {0, 3} are A1 and∅, {0}, {1}, {0, 1} are B1 soA1 = B1 = 4. Therefore T1 = 8, and Tn+1= 6Tn+ 2Bn = 7Tnforn > 1, hence Tn= 8· 7n−1.
Remarks. The problem becomes more challenging if we work with remain-ders modulo4n rather than with their set of representants. In this case the result is 7n, but writing recurrence relations is far from being obvious. Counting a related one-to-one set of differently defined subsets will do the trick.
Problem 16. Letp, q be two integers, q > p > 0. Let n > 2 be an integer and a0= 0, a1>0, a2,. . ., an−1,an= 1 be real numbers such that
ak6 ak−1+ ak+1
2 , k = 1, 2, . . . , n− 1.
Prove that
(p + 1)
n−1X
k=1
apk>(q + 1)
n−1X
k=1
aqk.
Solution. It immediately follows that
0 = a06a16· · · 6 an= 1 and
0 6 a1= a1− a06a2− a16· · · 6 an− an−1= 1− an−1.
A useful observation is that it suffices to prove the inequality forq = p + 1, as we may then extend it step-by-step top + 2, p + 3, . . . .
LetS(m, r) = Pm k=1
ark. By Abel’s summation formula we get
S(n, p + 1) = Xn k=1
ak· apk= anS(n, p)−
n−1X
k=1
(ak+1− ak)S(k, p).
Sincean = 1, this yields
S(n− 1, p + 1) = S(n − 1, p) −
n−1X
k=1
(ak+1− ak)S(k, p).
Now,aj− aj−16ak+1− ak, forj = 1, 2, . . . , k, so
(ak+1− ak)S(k, p) >
Xk j=1
(aj− aj−1)apj.
But
(aj− aj−1)apj >ap+1j − ap+1j−1 p + 1 , for
pap+1j + ap+1j−1>(p + 1)aj−1apj by the weighted AM-GM inequality.
Therefore
S(n− 1, p + 1) 6 S(n − 1, p) − 1 p + 1
n−1X
k=1
Xk j=1
(ap+1j − ap+1j−1)
= S(n− 1, p) − 1
p + 1S(n− 1, p + 1), where we useda0= 0 and the telescoping of the inner sum.
Consequently,
(p + 2)S(n− 1, p + 1) 6 (p + 1)S(n − 1, p).
Equality occurs forp = q, or a1=· · · = an−1= 0, or p = 0, q = 1 and ak= k n. Problem 17. Letk be a positive integer and n = 4k + 1. Let A ={a2+ nb2: a, b∈ Z}. Prove that there exist integers x, y such that xn+yn∈ A and x+y /∈ A.
Solution. It looks natural, in order to facilitate further factorizations, to take y = qx. Since squares are congruent to 0 or 1 modulo 4, and n≡ 1 (mod 4), any number of the forma2+nb2is congruent to 0, 1 or 2, but not 3, modulo 4. In order to havex+y /∈ A it is then enough to have x+y ≡ 3 (mod 4), that is, (q +1)x ≡ 3 (mod4), therefore we need take q even. Now, as n > 1, x = 1 + qn ≡ 1 (mod4), so we need take q + 1≡ 3 (mod 4), that is q ≡ 2 (mod 4). It remains to findx such that xn+ yn= (1 + qn)xnis of the forma2+ nb2.
Simply takingx = 1 + qnyieldsxn+ yn= (1 + qn)n+1=
(1 + qn)n+12 2
, of the forma2+ nb2, whereb = 0.
It is rather more difficult to find anx such that neither a = 0, nor b = 0. Take any integersu, v such that u2+v2≡ 1 (mod 4) and take x = (1+qn)(u2+nv2).
Thenx≡ 1 (mod 4), ensuring x + y /∈ A, and xn+ yn= (1 + qn)n+1(u2+ nv2)n
=
u(u2+ nv2)n−12 (1 + qn)n+12 2
+ n
v(u2+ nv2)n−12 (1 + qn)n+12 2
.
Finally, the requirements for the exhibited solution areq ≡ 2 (mod 4) and u + v≡ 1 (mod 2).
This result seems to have been a Sophie Germain conjecture.
Clearly, this method also works for any odd exponentm > 1 (instead of n), but not forn≡ 3 (mod 4).
Problem 18. Let m and n be positive integers and let S be a subset with (2m− 1)n + 1 elements of the set {1, 2, . . . , 2mn}. Prove that S contains m + 1 distinct numbersa0, a1, . . . , amsuch thatak−1| akfor allk = 1, 2, . . . , m.
Solution. We shall prove a stronger statement: such a setS will contain a subset{x, 2i1x, . . . , 2imx} with m + 1 elements (obviously fulfilling the original assertion). Assume this statement false and take among all sets for which it fails one withmin(S) maximal. As S has 2mn− (n − 1) elements, it follows S ∩ {1, 2, . . . , n} 6= ∅, whence 1 6 x = min(S) 6 n, so 2mx 6 2mn. Therefore, the set{x, 2x, . . . , 2mx} cannot be included in S, so there is some 1 6 i 6 m such that2ix /∈ S. Take S0 = (S∪ {2ix})\{x}; obviously min(S0) > min(S), hence there will exist a subsetA0 ={y, 2j1y, . . . , 2jmy} ⊂ S0. As this subset cannot be
Solution. We shall prove a stronger statement: such a setS will contain a subset{x, 2i1x, . . . , 2imx} with m + 1 elements (obviously fulfilling the original assertion). Assume this statement false and take among all sets for which it fails one withmin(S) maximal. As S has 2mn− (n − 1) elements, it follows S ∩ {1, 2, . . . , n} 6= ∅, whence 1 6 x = min(S) 6 n, so 2mx 6 2mn. Therefore, the set{x, 2x, . . . , 2mx} cannot be included in S, so there is some 1 6 i 6 m such that2ix /∈ S. Take S0 = (S∪ {2ix})\{x}; obviously min(S0) > min(S), hence there will exist a subsetA0 ={y, 2j1y, . . . , 2jmy} ⊂ S0. As this subset cannot be