# Problem 3.

In-1 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=1$ has two integer roots, and all these roots are less than (-1). Find the smallest value of $a$.
Answer: 15.

Solution. By Vieta's theorem, the product of the roots of the first equation is $a$, and the product of the roots of the second equation is $a-1$. Since the roots are integers and less than -1, their product is greater than 1, so each of the two consecutive numbers $a-1$ and $a$ is the product of two different integers greater than 1. Since the first odd number that is neither prime nor the square of a prime is 15, we get $a-1=14, a=15$. Then the roots of the first equation are -3 and -5 (in which case $b=8$), and the roots of the second equation are -2 and -7 (in which case $c=9$).

V-2 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=-1$ has two integer roots, and all these roots are greater than 1. Find the smallest value of $a$.

Answer: 14.

V-3 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=1$ and $x^{2}+c x+a=0$ has two integer roots, and all these roots are less than -1. Find the smallest value of $a$.

Answer: 15.

V-4 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=-1$ and $x^{2}+c x+a=0$ has two integer roots, and all these roots are greater than 1. Find the smallest value of $a$.

Answer: 14.

V-5 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=11, P(1)=21$?

Answer: 315.

Solution. Since $P(-1)=11$, we have $-1+A-B+C-D+E=11$; since $P(1)=21$, we have $1+A+B+C+D+E=21$. Therefore, $A+C+E=16, B+D=4$. The second of these two equations has 3 pairs of solutions in natural numbers, and the first equation has $C_{15}^{2}=\frac{15 \cdot 14}{2}=105$ solutions (arrange 16 "balls" in a row and divide them into three groups using two "partitions"). This corresponds to $3 \cdot 105=315$ different polynomials.

V-6 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=8, P(1)=22$?

Answer: 455.

V-7 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=9, P(1)=21$?

Answer: 364.

V-8 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=10, P(1)=22$?

Answer: 420.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final Stage 2020/21 Academic Year for 10-11 Grades

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