Solutions to the Shortlisted Problems of IMO 1992

4.33 Solutions to the Shortlisted Problems of IMO 1992

1. Assume that a pair (x, y) with x < y satisfies the required conditions. We claim that the pair (y, x 1 ) also satisfies the conditions, where x 1 = y 2 +m x (note that x 1 >y is a positive integer). This will imply the desired result, since starting from the

pair (1, 1) we can obtain arbitrarily many solutions. First, we show that gcd (x 1 , y) = 1. Suppose to the contrary that gcd(x 1 , y) = d >

1. Then d

2 |x 2 1 |y + m ⇒ d | m, which implies d | y | x + m ⇒ d | x. But this last is impossible, since gcd (x, y) = 1. Thus it remains to show that x 1 |y 2 +m

and y 2 |x 1 + m. The former relation is obvious. Since gcd(x, y) = 1, the latter is equivalent to y | (xx 1 ) 2 + mx 2 =y 4 + 2my 2 +m 2 + mx 2 , which is true because y | m(m + x 2 ) by the assumption. Hence (y, x 1 ) indeed satisfies all the required conditions.

Remark. The original problem asked to prove the existence of a pair (x, y) of positive integers satisfying the given conditions such that x + y ≤ m + 1. The problem in this formulation is trivial, since the pair x = y = 1 satisfies the con- ditions. Moreover, this is sometimes the only solution with x + y ≤ m + 1. For

example, for m = 3 the least nontrivial solution is (x 0 ,y 0 ) = (1, 4).

2. Let us define x n inductively as x n = f (x n −1 ), where x 0 ≥ 0 is a fixed real number. It follows from the given equation in f that x n +2 = −ax n +1 + b(a + b)x n . The general solution to this equation is of the form

n = λ 1 b n + λ 2 (−a − b) ,

where λ 1 , λ 2 ∈ R satisfy x 0 = λ 1 + λ 2 and x 1 = λ 1 b − λ 2 (a + b). In order to have x n ≥ 0 for all n we must have λ 2 = 0. Hence x 0 = λ 1 and f (x 0 )=x 1 = λ 1 b = bx 0 . Since x 0 was arbitrary, we conclude that f (x) = bx is the only possible solution of the functional equation. It is easily verified that this is indeed a solution.

3. Consider two squares AB ′ CD ′ and A ′ BC ′ D . Since AC ⊥ BD, these two squares are homothetic, which implies that the lines AA ′ , BB ′ ,CC ′ , DD ′ are concurrent at

a certain point O. Since the rotation about A by 90 ◦

takes ∆ ABK into ∆ AFD , it follows that

K BK ⊥ DF. Denote by T the intersec-

tion of BK and DF. The rotation about A some point X by 90 ◦ maps BK into DF

TL if and only if T X bisects an angle be- tween BK and DF. Therefore ∠FTA =

D ∠AT K = 45 ◦ . Moreover, the quadri-

B ′ C lateral BA ′

A DT is cyclic. Therefore ∠BTA ′ = BDA ′ = 45 ◦ and consequently that the points A , T, A ′ are collinear. It

follows that the point O lies on a bisector of ∠BT D and therefore the rotation R about O by 90 ◦ takes BK into DF. Analogously, R maps the lines CE , DG, AI

into AH , BJ,CL. Hence the quadrilateral P 1 Q 1 R 1 S 1 is the image of the quadrilat- eral P 2 Q 2 R 2 S 2 , and the result follows.

564 4 Solutions

4. There are 36 possible edges in total. If not more than 3 edges are left undrawn, then we can choose 6 of the given 9 points no two of which are connected by an undrawn edge. These 6 points together with the edges between them form a two-colored complete graph, and thus by a well-known result there exists at least one monochromatic triangle. It follows that n ≤ 33. In order to show that n = 33, we shall give an example of a graph with 32 edges that does not contain a monochromatic triangle. Let us start with a complete

graph C 5 with 5 vertices. Its edges can be colored in two colors so that there is no monochromatic triangle ( Fig. 1 ). Furthermore, given a graph H with k vertices without monochromatic triangles, we can add to it a new vertex, join it to all vertices of H except A, and color each edge BX in the same way as AX . The obtained graph obviously contains no monochromatic triangles. Applying

this construction four times to the graph C 5 we arrive to an example like that shown on Fig. 2 .

Fig. 1 Fig. 2 Second solution. For simplicity, we call the colors red and blue.

Let r (k, l) be the least positive integer r such that each complete r-graph whose edges are colored in red and blue contains either a complete red k-graph or a complete blue l-graph. Also, let t (n, k) be the greatest possible number of edges in a graph with n vertices that does not contain a complete k-graph. These num- bers exist by the theorems of Ramsey and Turán. Let us assume that r (k, l) < n. Every graph with n vertices and t(n, r(k, l)) +1 edges contains a complete subgraph with r (k, l) vertices, and this subgraph con- tains either a red complete k-graph or a blue complete l-graph. We claim that t (n, r(k, l)) + 1 is the smallest number of edges with the above property. By the definition of r (k, l) there exists a coloring of the complete graph

H with r (k, l) − 1 vertices in two colors such that no red complete k-graph or blue complete l-graph exists. Let c ij

be the color in which the edge (i, j) of H is colored, 1 ≤ i < j ≤ r(k,l) − 1. Consider a complete r(k,l) − 1-partite graph

G with n vertices and exactly t (n, r(k, l)) edges and denote its partitions by P i ,

i = 1, . . . , r(k, l) − 1. If we color each edge of H between P i and P j (j < i) in the color c ij , we obviously obtain a graph with n vertices and t (n, r(k, l)) edges in two colors that contains neither a red complete k-graph nor a blue complete l -graph. Therefore the answer to our problem is t (9, r(3, 3)) + 1 = t(9, 6) + 1 = 33.

4.33 Shortlisted Problems 1992 565

5. Denote by K , L, M, and N the midpoints of the sides AB, BC,CD, and DA, re- spectively. The quadrilateral KLMN is a rhombus. We shall prove that O 1 O 3 k KM . Similarly, O 2 O 4 k LN, and the desired result follows immediately.

We have −−−→ O 1 O 3 = −−→ KM + O 1 K + MO −−→ 3 . Assume that ABCD is positively ori- ented. A rotational homothety R with angle −90 ◦

√ and coefficient 1 /

3 takes the vectors −→ BK and CM −→ into −−→ O 1 K and −−→ MO 3 respectively. Therefore

−−−→ O O KM −−→

Since LN ⊥ KM, it follows that R(LN) is parallel to KM and so is O 1 O 3 .

6. It is easy to see that f is injective and surjective. From f (x 2 + f (y)) = f ((−x) 2 +

f (y)) it follows that f (x) 2 = ( f (−x)) 2 , which implies f (−x) = − f (x) be- cause f is injective. Furthermore, there exists z ∈ R such that f (z) = 0. From

f (−z) = − f (z) = 0 we deduce that z = 0. Now we have f (x 2 ) = f (x 2 + f (0)) =

, and consequently f (x) = f ( x ) 2 > 0 for all x > 0. It also follows that f (x) < 0 for x < 0. In other words, f preserves sign. Now setting x > 0 and y = − f (x) in the given functional equation we obtain

0 + ( f (x)) 2 = f (x) 2

f (x − f (x)) = f ( x + f (−x)) = −x + f ( x ) 2 = −(x − f (x)). But since f preserves sign, this implies that f (x) = x for x > 0. Moreover, since

f (−x) = − f (x), it follows that f (x) = x for all x. It is easily verified that this is indeed a solution.

7. Let G 1 ,G 2 touch the chord BC at P , Q and touch the circle G at R, S respect- ively. Let D be the midpoint of the complementary arc BC of G. The homo- thety centered at R mapping G 1 onto G also maps the line BC onto a tangent of G parallel to BC. It follows that this line touches G at point D, which is therefore the image of P under the homothety. Hence R , P, and D are collinear. Since ∠DBP = ∠DCB = ∠DRB, it follows that △DBP ∼ △DRB and conse-

quently that DP · DR = DB 2 . Similarly, points S , Q, D are collinear and satisfy DQ ·DS = DB 2 = DP · DR. Hence D lies on the radical axis of the circles G 1 and

G 2 , i.e., on their common tangent AW , which also implies that AW bisects the √ angle BAD. Furthermore, since DB = DC = DW = DP · DR, it follows from the lemma of (SL99-14) that W is the incenter of △ABC.

Remark. According to the third solution of (SL93-3), both PW and QW contain the incenter of △ABC, and the result is immediate. The problem can also be solved by inversion centered at W .

8. For simplicity, we shall write n instead of 1992. Lemma. There exists a tangent n-gon A 1 A 2 ...A n with sides A 1 A 2 =a 1 ,A 2 A 3 =

a 2 ,...,A n A 1 =a n if and only if the system x 1 +x 2 =a 1 ,x 2 +x 3 =a 2 ,,...,x n +x 1 =a n

566 4 Solutions

has a solution (x 1 ,...,x n ) in positive reals.

Proof. Suppose that such an n-gon A 1 A 2 ...A n exists. Let the side A i A i +1 touch the inscribed circle at point P i (where A n +1 =A 1 ). Then x 1 =A 1 P n =A 1 P 1 , x 2 =A 2 P 1 =A 2 P 2 , ...,x n =A n P n −1 =A n P n is clearly a positive solution of the system (1). Now suppose that the system (1) has a positive real solution (x 1 ,...,x n ). Let us draw a polygonal line A 1 A 2 ...A n +1 touching a circle of radius r at points P 1 ,P 2 ,...,P n respectively such that A 1 P 1 =A n +1 P n =x 1 and

A i P i =A i P i −1 =x i for i = 2, . . . , n. Observe that

OA 1 = OA n +1 = x 2 1 +r 2

A A 1 n +1 ∠A 2 OA 3 + ··· + ∠A n OA n +1 =2·

and the function f (r) = ∠A 1 OA 2 +

r + ··· + arctan x n

(arctan x 1 ) is con-

A 2 O closed simple polygonal line if and

tinuous. Thus A 1 A 2 ...A n +1 is a

P 2 P n −1 only if f (r) = 360 ◦ . But such an r

A n −1 exists, since f (r) → 0 when r → ∞,

and f (r) → ∞ when r → 0. This proves the second direction of the lemma.

For n = 4k, the system (1) is solvable in positive reals if a i = i for i ≡ 1,2 (mod 4), a i = i + 1 for i ≡ 3 and a i = i − 1 for i ≡ 0 (mod 4). Indeed, one solution is given by x i = 1/2 for i ≡ 1, x i = 3/2 for i ≡ 3 and x i = i − 3/2 for i ≡ 0,2 (mod 4).

Remark. For n = 4k + 2 there is no such n-gon. In fact, solvability of the system (1) implies a 1 +a 3 + ··· = a 2 +a 4 + ···, while in the case n = 4k + 2 the sum

a 1 +a 2 + ··· + a n is odd. √

9. Since the equation x 3 − x − c = 0 has only one real root for every c > 2/(3 3 ), α is the unique real root of x 3 − x − 33 1992 = 0. Hence f n ( α )=f( α )= α .

Remark. Consider any irreducible polynomial g (x) in the place of x 3 −x−

33 1992 . The problem amounts to proving that if α and f ( α ) are roots of g, then any f (n) ( α ) is also a root of g. In fact, since g( f (x)) vanishes at x = α , it must be divisible by the minimal polynomial of α , that is, g (x). It follows by induction that g (f (n)

(x)) is divisible by g(x) for all n ∈ N, and hence g( f (n) ( α )) = 0.

10. Let us set S (x) = {(y,z) | (x,y,z) ∈ V }, S y (x) = {z | (x,z) ∈ S y } and S z (x) = {y | (x, y) ∈ S z }. Clearly S(x) ⊂ S x and S (x) ⊂ S y (x) × S z (x). It follows that

|V | = ∑ |S(x)| ≤ ∑ |S x ||S y (x)||S z (x)|

= |S x | ∑ |S y (x)||S z (x)|.

4.33 Shortlisted Problems 1992 567 Using the Cauchy–Schwarz inequality we also get

∑ |S y (x)||S z (x)| ≤ ∑ |S y (x)| ∑ |S z (x)| = |S y ||S z |.

Now (1) and (2) together yield |V | ≤ |S x ||S y ||S z |.

11. Let I be the incenter of △ABC. Since 90 ◦ + α /2 = ∠BIC = ∠DIE = 138 ◦ , we obtain that ∠A = 96 ◦ .

AE

B E ′ D ′ C Let D ′ and E ′

be the points symmetric to D and E with respect to CE and BD re- spectively, and let S be the intersection point of ED ′ and BD. Then ∠BDE ′ = 24 ◦ and ∠D ′ DE ′ = ∠D ′ DE − ∠E ′ DE = 24 ◦ ,which means that DE ′ bisects the angle SDD ′ . Moreover, ∠E ′ SB = ∠ESB = ∠EDS + ∠DES = 60 ◦ and hence SE ′ bisects the angle D ′ SB . It follows that E ′ is the excenter of △D ′ DS and

consequently ∠D ′ DC = ∠DD ′ C = ∠SD ′ E ′ = (180 ◦ − 72 ◦ )/2 = 54 ◦ . Finally, ∠C = 180 ◦ − 2 · 54 ◦ = 72 ◦ and ∠B = 12 ◦ .

12. Let us set deg f = n and deg g = m. We shall prove the result by induction on n. If n < m, then deg x [ f (x) − f (y)] < deg x [g(x) − g(y)], which implies that f (x) −

f (y) = 0, i.e., that f is constant. The statement trivially holds. Assume now that n ≥ m. Transition to f 1 (x) = f (x) − f (0) and g 1 (x) = g(x) −

g (0) allows us to suppose that f (0) = g(0) = 0. Then the given condition for y = 0 gives us f (x) = f 1 (x)g(x), where f 1 (x) = a(x, 0) and deg f 1 = n − m. We now have

a (x, y)(g(x) − g(y)) = f (x) − f (y) = f 1 (x)g(x) − f 1 (y)g(y) =[f 1 (x) − f 1 (y)]g(x) + f 1 (y)[g(x) − g(y)].

Since g (x) is relatively prime to g(x) − g(y), it follows that f 1 (x) − f 1 (y) =

b (x, y)(g(x) − g(y)) for some polynomial b(x,y). By the induction hypothe- sis there exists a polynomial h 1 such that f 1 (x) = h 1 (g(x)) and consequently

f (x) = g(x) · h 1 (g(x)) = h(g(x)) for h(t) = th 1 (t). Thus the induction is com- plete.

13. Let us define

(pqr − 1)

F (p, q, r) = (p − 1)(q − 1)(r − 1)

p −1 q −1 r −1

+ . (p − 1)(q − 1) (q − 1)(r − 1) (r − 1)(p − 1)

568 4 Solutions Obviously F is a decreasing function of p , q, r. Suppose that 1 < p < q < r are

integers for which F (p, q, r) is an integer. Observe that p, q, r are either all even or all odd. Indeed, if for example p is odd and q is even, then pqr − 1 is odd while (p − 1)(q − 1)(r − 1) is even, which is impossible. Also, if p,q,r are even then F (p, q, r) is odd. If p ≥ 4, then 1 < F(p,q,r) ≤ F(4,6,8) = 191/105 < 2, which is impossible. Hence p ≤ 3. Let p = 2. Then q, r are even and 1 < F(2, q, r) ≤ F(2,4,6) = 47/15 < 4. There- fore F (2, q, r) = 3. This equality reduces to (q − 3)(r − 3) = 5, with the unique solution q = 4, r = 8. Let p = 3. Then q, r are odd and 1 < F(3, q, r) ≤ F(3,5,7) = 104/48 < 3. There- fore F (3, q, r) = 2. This equality reduces to (q − 4)(r − 4) = 11, which leads to q = 5, r = 15. Hence the only solutions (p, q, r) of the problem are (2, 4, 8) and (3, 5, 15).

14. We see that x 1 =2 0 . Suppose that for some m , r ∈ N we have x m =2 r . Then inductively x m +i =2 r −i (2i + 1) for i = 1, 2, . . . , r and x m +r+1 =2 r +1 . Since every natural number can be uniquely represented as the product of an odd number and

a power of two, we conclude that every natural number occurs in our sequence exactly once. Moreover, it follows that 2k −1=x k (k+1)/2 . Thus x n = 1992 = 2 3 · 249 implies that x n +3 = 255 = 2 · 128 − 1 = x 128 ·129/2 =x 8256 . Hence n = 8253.

15. The result follows from the following lemma by taking n = 1992 ·1993 2 and M = {d,2d,...,1992d}. Lemma. For every n ∈ N there exists a natural number d such that all the

numbers d , 2d, . . . , nd are of the form m k (m , k ∈ N, k ≥ 2). Proof. Let p 1 ,p 2 ,...,p n

be distinct prime numbers. We shall find d in the form

d =2 α 2 3 α 3 ···n α n , where α i ≥ 0 are integers such that kd is a perfect p k th power. It is sufficient to find α i ,i = 2, 3, . . . , n, such that α i ≡ 0 (mod p j ) if i 6= j and α i ≡ −1 (mod p j ) if i = j. But the existence of such α i ’s is an immediate consequence of the Chinese remainder theorem.

16. Observe that x 4 +x 3 +x 2 + x + 1 = (x 2 + 3x + 1) 2 − 5x(x + 1) 2 . Thus for x =5 25 we have

N =x 4 +x 3 +x 2 +x+1

= (x 2 + 3x + 1 − 5 13 (x + 1))(x 2 + 3x + 1 + 5 13 (x + 1)) = A · B. Clearly, both A and B are positive integers greater than 1.

17. (a) Let n =∑ k i =1 2 a i , so that α (n) = k. Then

n 2 = ∑ 2 2a i + ∑ 2 a i +a j +1

i <j

has at most k (k+1) + k

2 = k 2 binary ones.

(b) The above inequality is an equality for all numbers n k =2 k .

4.33 Shortlisted Problems 1992 569 (c) Put n m

m =2 2 m −1 −∑ m =1 2 2 −2 j j , where m > 1. It is easy to see that α (n m )=

2 m − m. On the other hand, squaring and simplifying yields n 2

m =1+ ∑ <j 2 2 +1−2 −2 . Hence α (n i 2 m )=1+ m (m+1) 2 and thus

m +1 ij

α (n 2 m ) 2 + m(m + 1)

→ 0 as m → ∞.

α (n m )

2 (2 m

− m)

Solution to the alternative parts.

(d) Let n i +2 j +1 =∑ 2 2 +1

i =1 2 . Then n =∑ i =1 2 +∑ i <j 2 2 has exactly

k (k+1)

binary ones, and therefore α(n 2 α(n) ) = 2k k (k+1) → ∞.

(e) Consider the sequence n i constructed in part (c). Let θ > 1 be a con- stant to be chosen later, and let N i =2 m i n i − 1 where m i > α (n i ) is such that m i / α (n i )→ θ as i → ∞. Then α (N i )= α (n i )+m i − 1, whereas

N 2 =2 2m i i n 2 i −2 m i +1 n i + 1 and α (N 2 i )= α (n 2 i )− α (n i )+m i . It follows that

α (N 2 α

(n 2 i )+( − 1) α (n i ) θ −1

lim

= lim

i →∞ α (N i ) i →∞

(1 + θ ) α (n i )

θ +1 which is equal to γ ∈ [0,1] for θ 1 = +γ 1 −γ (for γ = 1 we set m i / α (n i ) → ∞). (f) Let be given a sequence (n i ) ∞ i =1 with α (n 2 i )/ α (n i )→ γ . Taking m i > α (n i )

and N i =2 m i n i + 1 we easily find that α (N i )= α (n i ) + 1 and α (N 2 i )= α (n 2 i )+ α (n i )+ 1. Hence α (N 2 i )/ α (N i )= γ + 1. Continuing this procedure we can construct a sequence t i such that α (t 2 i )/ α (t i )= γ +k for an arbitrary k ∈ N.

18. Let us define inductively f 1 (x) = f (x) = 1 x n +1 and f (x) = f ( f n −1 (x)), and let

g (x) = x + f (x) + f n 2 (x) + ··· + f n (x). We shall prove first the following state- ment. Lemma. The function g n (x) is strictly increasing on [0, 1], and g n −1 (1) =

F 1 /F 2 +F 2 /F 3 + ··· + F n /F n +1 . Proof. Since f

y (x) − f (y) = −x (1+x)(1+y) is smaller in absolute value than x − y, it follows that x > y implies f 2k (x) > f 2k (y) and f 2k +1 (x) < f 2k +1 (y), and moreover that for every integer k ≥ 0,

[f 2k (x) − f 2k (y)] + [ f 2k +1 2k (x) − f +1 (y)] > 0. Hence if x > y, we have g n (x) − g n (y) = (x − y) + [ f (x) − f (y)] + ··· +

[f n (x) − f n (y)] > 0, which yields the first part of the lemma. The second part follows by simple induction, since f k (1) = F k +1 /F k +2 .

If some x i = 0 and consequently x j = 0 for all j ≥ i, then the problem reduces to the problem with i − 1 instead of n. Thus we may assume that all x 1 ,...,x n are different from 0. If we write a i = [1/x i ], then x i = 1 a i +x i +1 . Thus we can regard x i

as functions of x n depending on a 1 ,...,a n −1 .

Suppose that x n ,a n −1 ,...,a 3 ,a 2 are fixed. Then x 2 ,x 3 ,...,x n are all fixed, and x 1 = 1 a 1 +x 2 is maximal when a 1 = 1. Hence the sum S = x 1 +x 2 + ··· + x n is maximized for a 1 = 1.

570 4 Solutions We shall show by induction on i that S is maximized for a 1 =a 2 = ··· = a i = 1.

In fact, assuming that the statement holds for i − 1 and thus a 1 = ··· = a i −1 = 1, having x n ,a n −1 ,...,a i +1 fixed we have that x n ,...,x i +1 are also fixed, and that x i −1 = f (x i ), . . . , x 1 =f i −1 (x i ). Hence by the lemma, S = g i −1 (x i )+x i +1 + ··· + x n is maximal when x i = 1 a i +x i +1 is maximal, that is, for a i = 1. Thus the induction is complete. It follows that x 1 + ··· + x n is maximal when a 1 = ··· = a n −1 = 1, so that x 1 + ··· + x n =g n −1 (x 1 ). By the lemma, the latter does not exceed g n −1 (1). This completes the proof.

Remark. The upper bound is the best possible, because it is approached by taking x

1 n close to 1 and inductively (in reverse) defining x i −1 = 1 1 +x i = a i +x i .

√ + 2(1 − 2 )x √ 2 √ √ +3+

19. Observe that f (x) = (x 4 + 2x 2 + 3) 2 − 8(x 2 − 1) 2 = [x 4

2 2 ][x 4 + 2(1 + 2 )x 2 +3−2 2 ]. Now it is easy to find that the roots of f are √ 4 √ 4

x 1 ,2,3,4 = ±i i 2 ±1

and x 5 ,6,7,8 = ±i

2 ±1 . In other words, x k = α i + β j , where α 2 i = −1 and β 4 j = 2.

We claim that any root of f can be obtained from any other using rational func- tions. In fact, we have

x 3 =− α i −3 β j +3 α i β 2 j + β 3 j ,

x 5 = 11 α i +7 β j − 10 α i β 2 j − 10 β 3 j x 7 = −71 α i − 49 β j + 35 α i β 2 j 3 + 37 β j ,

from which we easily obtain that α = 24 −1 (127x + 5x 3 i + 19x 5 + 5x 7 ), β j = 24 −1 (151x + 5x 3 + 19x 5 + 5x 7 ). Since all other values of α and β can be obtained as rational functions of α i and

β j , it follows that all the roots x l are rational functions of a particular root x k .

We now note that if x 1 is an integer such that f (x 1 ) is divisible by p, then p > 3 and x 1 ∈Z p is a root of the polynomial f . By the previous consideration, all remaining roots x 2 ,...,x 8 of f over the field Z p are rational functions of x 1 , since 24 is invertible in Z p . Then f (x) factors as

f (x) = (x − x 1 )(x − x 2 ) ··· (x − x 8 ), and the result follows.

20. Denote by U the point of tangency of the circle C and the line l. Let X and U ′ be the points symmetric to U with respect to S and M respectively; these points do not depend on the choice of P. Also, let C ′

be the excircle

4.33 Shortlisted Problems 1992 571 of △PQR corresponding to P, S ′ the

center of C ′ , and W ,W ′ the points of X tangency of C and C ′ with the line PQ respectively. Obviously,

S △W ′ S ′ P . Since SX kS ′ U ′ and SX :

△W SP ∼ W

′ W ′ = SP : S ′ = SW : S R P , we de- duce that ∆ SX P ∼ ∆ S ′ U ′ P , and conse-

quently that P lies on the line XU ′ . On the other hand, it is easy to show that

each point P of the ray U ′ X over X sat- W ′

isfies the required condition. Thus the desired locus is the extension of U ′

S X ′ over X .

21. (a) Representing n 2 as a sum of n 2 − 13 squares is equivalent to representing

13 as a sum of numbers of the form x 2 − 1, x ∈ N, such as 0,3,8,15,.... But it is easy to check that this is impossible, and hence s (n) ≤ n 2 − 14. (b) Let us prove that s (13) = 13 2 − 14 = 155. Observe that

Given any representation of n 2 as a sum of m squares one of which is even, we can construct a representation as a sum of m + 3 squares by dividing the even square into four equal squares. Thus the first equality enables us to construct representations with 5 , 8, 11, . . ., 155 squares, the second to con- struct ones with 7 , 10, 13, . . ., 154 squares, and the third with 9, 12, . . . , 153

squares. It remains only to represent 13 2 as a sum of k = 2, 3, 4, 6 squares. This can be done as follows:

= 11 2 +4 2 +4 2 +4 2 = 12 2 +3 2 +2 2 +2 2 +2 2 +2 2 . (c) We shall prove that whenever s (n) = n 2 − 14 for some n ≥ 13, it also holds

that s (2n) = (2n) 2 − 14. This will imply that s(n) = n 2 − 14 for any n =

2 t · 13. If n 2 =x 2 1 + ··· + x 2 r , then we have

(2n) 2 = (2x 1 ) 2 + ··· + (2x r ) 2 . Replacing (2x i ) 2 with x 2 i +x 2 i +x 2 i +x 2 i as long as it is possible we can

obtain representations of (2n) 2 consisting of r , r + 3, . . . , 4r squares. This gives representations of (2n) 2 into k squares for any k

≤ 4n 2 − 62. Further, we observe that each number m ≥ 14 can be written as a sum of k ≥ m

numbers of the form x 2 − 1, x ∈ N, which is easy to verify. Therefore if

572 4 Solutions k ≤ 4n 2 − 14, it follows that 4n 2 − k is a sum of k numbers of the form

x 2 − 1 (since k ≥ 4n 2 − k ≥ 14), and consequently 4n 2 is a sum of k squares. Remark. One can find exactly the value of s (n) for each n: 

 1 , if n has no prime divisor congruent to 1 mod 4; s (n) = 2 ,

if n is of the form 5 ·2 k , k a positive integer;  n 2 − 14, otherwise.

4.34 Shortlisted Problems 1993 573