# A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is: $P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to : $P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$ $P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$ $P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$ $P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

# Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff $f^{-1}(f(A))=A$ for all $A \subset X$. Given that $f: X \longrightarrow Y$.

Problem 2:

Prove that a function if is onto iff $f(f^{-1}(C))=C$ for all $C \subset Y$. Given that $f: X \longrightarrow Y$.

Problem 3:

(a) How many functions are there from a non-empty set S into $\phi$\?

(b) How many functions are there from $\phi$ into an arbitrary set $S$?

(c) Show that the notation $\{ X_{i} \}_{i \in I}$ implicitly involves the notion of a function.

Problem 4:

Let $f: X \longrightarrow Y$ be a function, let $A \subset X$, $B \subset X$, $C \subset Y$ and $D \subset Y$. Prove that

i) $f(A \bigcap B) \subset f(A) \bigcap f(B)$

ii) $f^{-1}(f(A)) \supset A$

iii) $f(f^{-1}(C)) \subset C$

Problem 5:

Let I be a non-empty set and for each $i \in I$, let $X_{i}$ be a set. Prove that

(a) for any set B, we have $B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})$

(b) if each $X_{i}$ is a subset of a given set S, then $(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$ where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) $A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)$

(ii) $(A-B) \times C=(A \times C)-(B \times C)$

🙂 🙂 🙂

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if $A \bigcup B=A$ and $A \bigcap B=A$, then $A=B$.

Problem 2:

Show that in general $(A-B)\bigcup B \neq A$.

Problem 3:

Let $A = \{ 2,4, \ldots, 2n, \ldots\}$ and $B= \{ 3,6,\ldots, 3n, \ldots\}$. Find $A \bigcap B$ and $A - B$.

Problem 4:

Prove that (a) $(A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)$

Prove that (b) $A \Delta B = (A \bigcup B)-(A \bigcap B)$

Problem 5:

Prove that $\bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})$

Problem 6:

Let $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\bigcup_{n=2}^{\infty}A_{n}$ (ii) $\bigcap_{n=2}^{\infty}A_{n}$.

Problem 7:

Find (i) $\bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}]$ (ii) $\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$

Problem 8:

Let $A_{\alpha}$ be the set of points lying on the curve $y=\frac{1}{x^{\alpha}}$ where $(0. What is $\bigcap_{\alpha \geq 1}A_{\alpha}$?

Problem 9:

Let $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\frac{1}{4} \leq y \leq \frac{3}{4}$? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let $\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \in M$, and let $aRb$ if and only if $(a,b) \in \mathscr{R}$. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

• Reflexive and symmetric, but not transitive.
• Reflexive, but neither symmetric nor transitive.
• Symmetric, but neither reflexive nor transitive.
• Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

# A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of $\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients: $\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as $\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get: $\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that $\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes: $(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.