This is an optional module for the students preparing for engineering entrance examinations
Q-1) Find the number of natural numbers \(n\) which satisfy the following 2 conditions:
a) \(n \le 1991\)
b) \(6\) divides \(n^2+3n+2\)
Solution:
\(n^2+3n+2=(n+1)(n+2)\), this is divisible by \(2\) for any natural number 2. Note that for any number to be divisible by \(6\), it must be divisible by its coprime factors which are \(23\). For \((n+1)(n+2)\) to be divisible by \(3\), let’s define the notion of congruences :
We write \(a \equiv b(\text{mod }n)\) \(\iff n|(a-b)\). For now, it’s recomended to go through this link to quickly grasp the concepts. We’ll cover this in more detail later on in our course.
So if \(n \equiv 1,2(\text{mod }3)\), we have \(3|(n+2)(n+1)\). Thus the total number of numbers divisible by 3 and less than \(1991\) is \(\frac{1989}{3} = 663\). Therefore the total number of numbers = \(1991-663=1328\).
Q-2) Solve the following system of equations in reals
\(\begin{Bmatrix}
x+y-z=4 & \\
x^2-y^2+z^2=-4 & \\
xyz=6 &
\end{Bmatrix}\)
Solution:
From the second equation, we can rearrange to have \((x-z)^2+2xz=y^2-4\). Using the other 2 equations we have \((4-y)^2+\frac{12}{y}=y^2-4 \Rightarrow 16-8y+\frac{12}{y}=-4\). Multiplying by \(y\) and solving the quadratic gives \(y= \left(3, -\frac{1}{2} \right)\). So for these 2 values of \(y\) we have \(z=\frac{2}{x}, -\frac{12}{x}\). Putting these values in the first equation and solving the quadratic gives \(x=-1,2\). Thus the set of values is \((x,y,z)=\left \{ (-1,3,-2), (2,3,1) \right \}\).
Q-3) Let \(f\) be a polynomial with integer coefficients such that for five distinct integers \(a_1,a_2,..a_5\) \(f\) takes the value \(2\). Show that there does not exist an integer \(b\) such that \(f(b)=9\)
Solution:
Let’s consider \(g(x)=f(x)-2\). \(g(x)\) has minimum 5 roots, thus \(g(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x)=f(x)-2\), where \(q\) is a polynomial with integer coefficients. If there exists such a \(b\), for which \(f(b)=9\), we have
But given that \(a_i\) are distinct and \(7\) can have only 2 distinct factors, this is impossible. Thus no such \(b\) exists.
Q-4) Given a monic polynomial \(f(x)\) of degree \(n\) with integer coefficients, with \(k,p \in \mathbb{N}\), prove that if none of the numbers \(f(k), f(k+1),..f(k+p)\) is divisible by \(p+1\), then \(f(x)=0\) has no rational solution.
Solution:
If a rational root exists for \(f\), then it must be an integer. Suppose that an integer root \(m\) exists. Then \(f(x)=(x-m)g(x)\), where \(g\) is a polynomial with integer coefficients. If we substitute \(x=k,k+1,…k+p\) we get \(f(k)=(k-m)g(k), \thinspace f(k+1)=(k+1-m)g(k+1),…f(k+p)=(k-p-m)g(k+p)\). Note that one of the \(p+1\) successive integers \(k-m,…k+p-m\) should be divisible by \(p+1\). Thus no rational root exists.
Q-5) Solve the following system of equations for reals \(a,b,c,d,e\)
\(\begin{Bmatrix}
3a=(b+c+d)^3\\
3b=(c+d+e)^3\\
3c=(d+e+a)^3\\
3d=(e+a+b)^3\\
3e=(a+b+c)^3
\end{Bmatrix}\)
Solution:
If we consider \(a\) to be the greatest of the 5 reals, we have \(d+e+a\ge c+d+e \Rightarrow c=\frac{(d+e+a)^3}{3}\ge \frac{(c+d+e)^3}{3}=b\). Also we have
\(e=\frac{(a+b+c)^3}{3}\ge \frac{(b+c+d)^3}{3}=a\). We already have \(a \ge e\), so this shows \(a=e\). This means \(e \ge b\).
\(b=\frac{(c+d+e)^3}{3} \ge \frac{(b+c+d)^3}{3}=a\thinspace \Rightarrow a=b\).
Similarly we have \(a=c\) & \(a=d\). Thus \(3a=27a^3\), and the solutions are
\(\left \{ (0,..0), \left(\frac{1}{3},..\frac{1}{3} \right ), \left(-\frac{1}{3},..-\frac{1}{3} \right ) \right \}\).
Q-6) Suppose \(a,b,c\) are 3 real numbers such that the quadratic equation \(x^2-(a+b+c)x+(ab+bc+ca)=0\) has roots of the form \(\alpha + i\beta\), where \(\alpha>0\) & \(\beta \ne 0\) are real numbers.
Show that:
a) the numbers \(a,b,c\) are all positive
b) the numbers \(\sqrt{a}, \sqrt{b}, \sqrt{c}\) form the sides of a triangle.
Solution:
We know that \(a+b+c>0\) and \((a+b+c)^2<4(ab+bc+ca)\). From this we have \((a+b-c)^2<4ab \thinspace \Rightarrow ab>0\). Similarly we have \(bc>0\) and \(ca>0\). Either we have all 3 \(a,b,c\) negative or all of them as positive, note that \(a+b+c>0\) which implies that all of them are positive.
For the second part, let’s assume \(a<b<c\thinspace \Rightarrow \sqrt{a}\sqrt{b}<\sqrt{c}\), thus it suffices to show that \(\sqrt{a}+\sqrt{b}>\sqrt{c}\). Assume the opposite, thus we have \(\sqrt{a}+\sqrt{b}<\sqrt{c}\thinspace \Rightarrow a+b+2\sqrt{ab}<c \thinspace \Rightarrow a+b-c<-2\sqrt{ab}\). Note that squarring both sides gives \((a+b-c)^2>4ab\), which is opposite to what we have shown above.
Q-7) Show that there does not exist polynomials \(p,q\) with integer coefficients such that \(p(x)q(x)=x^5+2x+1\).
Solution:
This is the right time to expose Cohn’s irreducibility criteria. It states that for any polynomial \(f\), if \(f(x)\) is a prime for some \(x\ge 2\), then \(f\) is irreducible over \(\mathbb{Z}\), which simply means that it cannot be factored into polynomials with integer coefficients. Let’s try with \(x=4\) in our example. We have \(x^5+2x+1=1024+9=1033\), which is a prime number, and thus such polynomials \(p,q\) do not exist.
Q-8) Let \(p,q\) be 2 polynomials with equal sum of their coefficients. Show that there exists an integer \(a\) and a polynomial \(r(x)\), with \(r(1) \ne 0\) such that \(p(x)-r(x)=(x-1)^ar(x)\).
Solution:
\(p(x)-q(x)=(x-1)^ar(x)\thinspace \iff\) \(1\) is a root of \(p(x)-r(x)\). Note that we have \(p(1)=r(1)\), which means that 1 is a root of \(p-q\). Thus there exists some integer \(a\), which is the multiplicity of the root \(1\), and some polynomial \(r\) satisfying the given expression.
Q-9) Let \(a,b,c\) be 3 reals such that \(1\ge a\ge b\ge c \ge 0\). Show that if \(\lambda\) be a root of the equation \(x^3+ax^2+bx+c=0\) (real or non real), then \(|\lambda| \le 1\).
Solution:
If \(\lambda\) be a root then we have
Adding the 2 we have
And from triangle inequality it follows
Assume that \(|\lambda|\ge 1\).
Thus \(|\lambda|^4 \ge|\lambda|^3= |\lambda|^3(1-a+a-b+b-c+c)=|\lambda|^3(1-a)+|\lambda|^3(a-b)+|\lambda|^3(b-c)+|\lambda|^3c\).
Also, since \(|\lambda|>1\), we can conclude that
\(|\lambda|^4 \ge |\lambda|^3(1-a)+|\lambda|^3(a-b)+|\lambda|^3(b-c)+|\lambda|^3c \ge |\lambda|^3(1-a)+|\lambda|^2(a-b)+|\lambda|(b-c)+c\).
Combining this with what we have proved earlier shows that \(|\lambda| \le 1\)
Q-10) Show that the equation \(8x^4-16x^3+16x^2-8x+k=0\) has atleast one non real root. Find the sum of all non real roots.
Solution:
We start with the observation that all the coefficients of the non constant terms of the polynomial are powers of \(2\). Thus if we consider the polynomial \(8x^4-16x^3+16x^2-8x\), and put in \(x=\frac{1}{2}\), we observe that the value comes out to be \(-\frac{3}{2}\). Thus \(x=\frac{1}{2}\) should be a root of \(8x^4-16x^3+16x^2-8x+\frac{3}{2}\). Thus we express the given polynomial as \(f(x)=8x^4-16x^3+16x^2-8x+\frac{3}{2}+(k-\frac{3}{2})\), and we keep factorising it by the Remainder Theorem with \(x=\frac{1}{2}\).
This leads us to \(f(x)=\frac{1}{2}(2x-1)^2\left \{ (2x-1)^2+2 \right \}+k-\frac{3}{2}\). Note that for this polynomial, we have \(f\left(\frac{1}{2} +x\right) = f\left(\frac{1}{2}-x \right)\), which means that \(f(x)\) is symmetric about \(x=\frac{1}{2}\). Thus if there exists real roots, they’ll be of the form \(\frac{1}{2}+x_0\) & \(\frac{1}{2}-x_0\).
Also note that if \(\alpha_i\) denote the roots of this equation, then \(\sum \alpha_i^2=0\), which implies that since not all roots are \(0\), there exists non-real roots of the equation \(f(x)=0\), and from the above argument, sum of real roots = \(1\) if they exist.
Thus sum of non real roots = \(1\) if there exist real roots and \(2\) if real roots don’t exist.
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