A fifth degree equation in two variables: a clever solution


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:


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 :





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.

Nalin Pithwa.

A quadratic equation question for pRMO or preRMO


Find the necessary and sufficient condition that the quadratic equation ax^{2}+bx+c=0 where a \neq 0 has one root which is the square of the other.


Let the two roots of the given quadratic equation ax^{2}+bx+c=0, with a \neq 0 be \alpha and \beta such that \beta = \alpha^{2}.

Then, we know \alpha+\beta=-\frac{b}{a} and \alpha\beta=\frac{c}{a} so that \alpha+\alpha^{2}=-\frac{b}{a} and \alpha^{3}=\frac{c}{a}. From the latter relation, we get that \alpha = (\frac{c}{a})^{\frac{1}{3}}. Substituting this in the first relation of sum of roots, we get the following necessary and sufficient condition:

(\frac{c}{a})^{\frac{1}{3}} + (\frac{c}{a})^{\frac{2}{3}} = -\frac{b}{a}.

The above is the desired solution.

🙂 🙂 🙂

Nalin Pithwa

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


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)}


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


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


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.

preRMO or RMO algebra: my student’s solutions

(These are the solutions from a student of mine, whose identity is private. I will call him, RI, Bengaluru.)

Question I: 

Solve as elegantly as possible: t^{4}=49+20\sqrt{6}

Solution I (of RI, Bengaluru):

t^{4}=49+ 2\times 5 \times \sqrt{24}



t^{2}=5 + 2\sqrt{6}

Hence, t=\sqrt{2}+\sqrt{3}.

Question II: 

Find the necessary and sufficient conditions on the coefficients p, q, and r of the given cubic equation such that the roots of the cubic are in AP:


Solution II: (credit to RI, Bengaluru) : 

Let the roots of the above cubic be a-R, a, a+R.

By Viete’s relations: (a-R)+a +(a+R)=-p and a(a-R)+(a-R)(a+R)+(a+R)(a)=q and a(a-R)(a+R)=r so that we get 3a=p so that a=\frac{p}{3} and from the second equation we get a^{2}-aR+a^{2}-R^{2}+a^{2}+Ra=q, that is, 3a^{2}-R^{2}=q so that a^{2}-R^{2}=q-2\times a^{2} = q - 2 \times (\frac{p}{3})^{2}=q-\frac{2}{9}\times p^{2}, and exploiting the third Viete’s relation we get a \times (a^{2}-R^{2})=r, that is (q - (\frac{2}{9})\times p^{2})a=r, which is the required necessary and sufficient condition, where a=\frac{p}{3}.

Method II: For pedagogical purposes. The above solution to question 2 was quick and elegant because of the right choice of three quantities in AP as a-R, a, a+R. Do you want to know how ugly and messy it can get if a standard assumption is made:

Let the roots of the cubic be \alpha, \beta, \gamma. Let d be the common difference. So that the three roots in AP are \alpha, \beta=\alpha+d and \gamma=\alpha + 2d

Then, applying Viete’s relations, we get \alpha + \beta + \gamma = -p so that 3\alpha + 2d=-p and \alpha\beta+ \beta\gamma + \gamma\alpha=q which changes to \alpha \times (\alpha + d) + (\alpha+d)(\alpha+2d)+ (\alpha+d)(\alpha+2d) = q and the third viete’s relation gives us \alpha (\alpha+d)(\alpha+2d)=r.

The second Viete’s relation is a quadratic in \alpha and the third Viete’s relation is a cubic in \alpha. This is how messy it can get…at least, you will agree RI’s judicious choice has rendered a clean, quick solution.


Nalin Pithwa.