Download UI Putnam Training Sessions Problem Set 18: Polynomials, II

UI Putnam Training Sessions
Problem Set 18: Polynomials, II Solutions
http://www.math.illinois.edu/contests.html
The Problems
1. Congruences and divisibility of polynomials, II: Nonexistence of prime-producing polynomials.
Suppose P (x) is a nonconstant polynomial with integer coefficients. Prove that there exist infinitely
many positive integers n such that |P (n)| is composite. (As a consequence, there does not exist a
nonconstant polynomial that only produces prime values.)
To solve this, proceed as follows:
(a) First assume that P (x) has constant coefficient a0 with |a0 | ≥ 2. Prove that there are infinitely
many integers n such that P (n) is divisible by a0 (and hence composite). (Hint: The key is
to apply the congruence properties for polynomials with the “right” modulus m.)
(b) Next prove the assertion in the case when P (x) has constant coefficient a0 = 0.
(c) Finally, prove the assertion in the remaining case when P (x) has constant coefficient a0 = ±1.
(This is the hardest case.)
Solution: Since P (x) is nonconstant, we have |P (n)| → ∞ as n → ∞. Thus, there exists a positive
integer n0 such that |P (n0 )| ≥ 2. Let m0 = |P (n0 )|. Since P (x) has integer coefficients, m0 is an
integer. By the properties of congruences, we have P (n) ≡ P (n0 ) = ±m0 ≡ 0 mod m0 for any n
with n ≡ n0 mod m0 . Hence P (n) is divisible by m0 for any such n. Since m0 ≥ 2 and |P (n)| → ∞
as n → ∞, it follows that P (n) is composite for any sufficiently large n with n ≡ n0 mod m0 .
2. Roots and factors of a polynomial, II.
k
(a) (USAMO 1975) Let P (x) be a polynomial of degree n satisfying P (k) = k+1
for k = 0, 1, . . . , n.
Find P (n + 1). (Hint: Construct a polynomial with roots at 0, 1, . . . , n.)
Solution: Answer: 1 if n is odd, 1−1/(n+2) if n is even. For the proof, consider the polynomial
Q(x) = (x + 1)P (x) − x. Q(x) is a polynomial of degree n + 1 with roots at 0, 1, . . . , n, so
it must be of the form Q(x) = Cx(x − 1) . . . (x − n). To determine C, we use the fact that
Q(−1) = 0·P (−1)−(−1) = 1. Then, 1 = Q(−1) = C(−1)(−2) . . . (−n−1) = C(−1)n+1 (n+1)!,
so C = (−1)n+1 /(n + 1)!. Therefore Q(n + 1) = C(n + 1)n · · · 1 = C(n + 1)! = (−1)n+1 , and
P (n + 1) = (Q(n + 1) + (n + 1))/(n + 2) = 1 − ((−1)n+1 + 1)/(n + 2), which gives the above
answer.
(b) Let P (x) be a polynomial with integral coefficients. Suppose that there exist four distinct
integers ai (i = 1, . . . , 4) such that P (ai ) = 5 for i = 1, . . . , 4. Show that there exists no integer
k such that P (k) = 8.
Solution: Applying the Factor Theorem to the polynomial P (x) − 5, we get P (x) − 5 =
Q
4
with integer coefficients. Now if P (k) = 8 for
i=1 (x − ai )Q(x), where Q(x) is a polynomial
Q4
some integer k, we would have 8 − 5 = i=1 (k − ai )Q(k). Since Q(x) has integer coefficients,
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UI Putnam Training
Problem Set 18: Polynomials, II Solutions
Fall 2016
Q
Q(k) must be an integer. Hence 4i=1 (k − ai ) must divide 3. which is only possible if one of
the factors k − ai or Q(k) is equal to ±3 and the other three factors are ±1. But in this case
two of the factors must be equal, which contradicts the distinctness of the numbers ai .
(c) Let a, b, c be pairwise distinct integers. Prove that there does not exist a polynomial with
integer coefficients that cyclically permutes these values, i.e., that satisfies P (a) = b, P (b) = c,
and P (c) = a. (Hints: Try first a concrete case, e.g., with a = 1, b = 2, c = 3. Use congruences
or the factor theorem to show that there is no polynomial satisfying P (1) = 2, P (2) = 3, P (3) =
1. Then try to do the same with general values of a, b, c.)
Solution: Suppose that P (x) is a polynomial with the given property. Then a is a root of
P (x) − b, and so by the factor theorem P (x) − b = (x − a)Q(x), where Q(x) is a polynomial
with integer coefficients. In particular, the number (c − b)/(b − a) = (P (b) − b)/(b − a) = Q(b)
must be an integer. Similarly, each of the numbers (a − c)/(c − b) and (b − a)/(a − c) must
also be an integer. Multiplying these three numbers we get
c−b a−c b−a
·
·
= 1,
b−a c−b a−c
which is only possible if each of the three factors is 1 or −1, i.e., if |c − b| = |a − b| = |a − c|.
But the latter is clearly impossible for distinct integers a, b, c.
3. Roots and factors of a polynomial, III. Here is a final challenge, from the Putnam.
(B1, Putnam 1985) Let k be the smallest positive integer for which there exist distinct integers
m1 , m2 , m3 , m4 , m5 such that the polynomial
p(x) = (x − m1 )(x − m2 )(x − m3 )(x − m4 )(x − m5 )
has exactly k nonzero coefficients. Find, with proof, a set of integers m1 , m2 , m3 , m4 , m5 for which
this minimum k is achieved.
Solution: Taking the mi ’s to be 0, ±a, ±b, with a, b distinct positive integers, we get p(x) =
x(x2 − a2 )(x2 − b2 ), which has 3 nonzero coefficients. This shows that k = 3 works. It remains to
show that a smaller k-value doesn’t work. Since p(x) must have degree 5 and leading coefficient 1,
the only possibilities with 2 or 1 terms are x5 + ax4 , x5 + ax3 , x5 + ax2 , x5 + ax, x5 + a, x5 , but
none of these can have 5 distinct real roots. (The first three of these polynomials have multiple
roots at x = 0, while the last two have complex roots.)
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UI Putnam Training
Problem Set 18: Polynomials, II Solutions
Fall 2016
Challenge Problem of the Week: Universal primes
If a prime number is written out in decimal and then reinterpreted in a different
basis, the resulting number need not be a prime. For example, 23 is a prime in the
decimal system, but in base 11, it becomes (23)11 = 25, which is not a prime. On the
other hand, since 61, (61)11 = 67, (61)12 = 73, and (61)13 = 79 are all primes, the
number 61 is prime in base b = 10, and remains so in each of the bases b = 11, 12, 13
(though in the next basis, b = 14, it represents a composite number).
Call a prime number universal, if it remains prime with respect to all bases b > 10.
Clearly, any single-digit prime is universal since it represents the same number in all
such bases. Thus, the real question is:
Do there exist multi-digit universal primes? That is, do there exist digits
a1 , a2 , . . . , ak ∈ {0, 1, . . . , 9}, with k ≥ 2, a1 6= 0, whose concatenation,
interpreted as a base-b integer, (a1 a2 . . . ak )b , is a prime for all bases
b = 10, 11, 12, . . . ?
Solution: Note that, given a string a1 a2 . . . ak with “digits” ai , the integer represented by this
string in base b is just the value of the polynomial P (x) = a1 xk−1 + a2 xk−2 + · · · + ak at x = b. Further,
multi-digit integers correspond to non-constant polynomials in this way. Thus, the question whether
multi-digit “universal primes” exist reduces to the following question:
Are there non-constant polynomials P (x) with nonnegative integer coefficients such that P (x)
is prime for all sufficiently large integers x?
The answer to the latter question (and hence to the original question about the existence of multidigit universal primes) turns out to be “no”. This can be seen with a little bit of modular arithmetic as
follows:
Let n0 = P (2) and note that, since P (x) is non-constant (i.e., has degree at least 1) and has nonnegative integer coefficients, n0 is an integer ≥ 2. Now consider any integer n with n ≡ 2 mod n0 . By
the general properties of congruences, we have ni ≡ 2i mod n0 for any nonnegative integer i, and hence
P (n) = a1 nk−1 + a2 nk−2 + · · · + ak n0 ≡ a1 2k−1 + a2 2k−2 + · · · + ak 20 = P (2) mod n0 . Since P (2) = n0 ,
it follows that P (n) ≡ 0 mod n0 , so P (n) is divisible by n0 for any such n. Moreover, since P has nonnegative coefficients and is non-constant, P (x) is increasing in x and so P (n) > P (2) = n0 for any n > 2.
Thus, n0 is a proper divisor of P (n) for any integer n > 2 with n ≡ 2 mod n0 , and P (n) is therefore
composite for any such n, i.e., for n = 2 + n0 , 2 + 2n0 , . . . .
Happy Problemsolving!
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