RMO 2018 Tamil Nadu Problem 3 - Nonlinear Diophantine Equation

RMO 2018 Tamil Nadu Problem 3 is from Number Theory. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.

Problem

Show that there are infinitely many 4-tuples (a, b, c, d) of natural numbers such that $a^3 + b^4 + c^5 = d^7$.

Key ideas you will need to solve this problem

Also see

Advanced Math Olympiad Program

Hint 1: Powers of 3

Powers of 3 have a very interesting property: $$ 3^n + 3^n + 3^n = 3^{n+1} $$

This simple observation is the key to this problem.

Hint 2: Expressing \( 3^n \) in multiple ways.

We want $ a^3 = 3^n, b^4 = 3^n, c^5 = 3^n $. Clearly n needs to be a multiple of 3, 4 and 5. For example, $3 \times 4 \times 5 = 60 $ may work. That is $$ 3^{60} + 3^{60} + 3^{60} = (3^{20})^3 + (3^{15})^4 + (3^{12})^5 $$

Hence in this case $$ a = 3^{20}, b = 3^{15}, c = 3^{12} $$

This will work for any multiple of 60. Suppose 60k is a multiple of 60 (that is k is any integer). Then we will have $$ a = 3^{20k}, b = 3^{15k}, c = 3^{12k} $$

Hint 3: We need 60k +1 to be a multiple of 7

Notice that $$ 3^{60k} + 3^{60k} + 3^{60k} = (3^{20k})^3 + (3^{15k})^4 + (3^{12k})^5 = 3^{60k+1} $$

We need $$ 3^{60k+1}  = d^7 $$

That is 60k +1 needs to be a multiple of 7. In terms of modular arithmetic we want $$ 60k + 1 \equiv 0 \mod 7 $$

$60 \equiv 4 \mod 7 \  \Rightarrow 60k + 1 \equiv 4k +1 \mod 7 \ \Rightarrow 4k \equiv - 1 \equiv 6 \mod 7$

This is where we will use the notion of inverse of a number modulo 7. Inverse of 4 modulo 7 is 2. This is because $ 4 \times 2 = 8 \equiv 1 \mod 7 $. The Bezout's theorem guarantees existence of inverse of 4 modulo 7. (Look into the reference at the end of this discussion if you do not know these ideas).

$4k \equiv 6 \mod 7 \ \Rightarrow 2\times 4k \equiv 2\times 6 \mod 7 \ \Rightarrow k \equiv 5 mod 7 $

Hence k = 7k' + 5 is suitable for our purpose.

Since there are infinitely many such integers with have infinitely many 4 tuples that will work.

Illustration: For k' = 0, k = 5. Therefore $60 \times 5 = 300$ should work. And it does: $$ 3^{300} + 3^{300} + 3^{300} = (3^{100})^3 + (3^{75})^4 + (3^{60})^5 = 3^{301} = (3^{43})^7 $$

Reference:

Also see

RMO 2018 Tamil Nadu Region

RMO 2018 Tamil Nadu Problem 1 - angle bisector

RMO 2018 Tamil Nadu Problem 1 is from Geometry. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.

Problem

Let ABC be an acute-angled triangle and let D be an interior point of the line segment BC. Let the circumcircle of triangle ACD intersect AB at E (E between A and B) and let the circumcircle of triangle ABD intersect AC at F (F between A and C). Let O be the circumcenter triangle AEF. Prove that OD bisects ( \angle EDF )

Key ideas you will need to solve this problem

Also see

Advanced Math Olympiad Program

Hint 1: Draw a picture

RMO 2018 Tamil Nadu Problem 1

Hint 2: A construction would be nice

Join OE and OF. What kind of quadrilateral is OEDF? (Pause here. Try it yourself.)

RMO 2018 Tamil Nadu Problem 1 Solution

Hint 3: Some angle chasing

Note that $ EDCA$  is a cyclic quadrilateral (all of its four vertices are on a circle). Hence $ \angle ODB = \angle EAC = \angle A $ .

Similarly $AFDB $ is cyclic. Hence $\angle CDF = \angle FAB = \angle A $

This implies $ \angle EDF = 180^o - 2 \angle A $. Since O is the center of $ \Delta AEF $, $ \angle EOF = 2\times \angle EAF = 2 \times \angle A$.

Therefore $ \angle EDF + \angle EAF = 180^o - 2 \times \angle A + 2 \times \angle A = 180^o $.  This implies quadrilateral OEDF is cyclic.

Hint 4: OE = OF

OE = OF because O is the center and E, F are at the circumference of the circle passing through AEF. Therefore both are radii of the same circle.

This implies $ \angle OEF = \angle OFE $

Hint 5: Final Lap; Some more angle chasing

Since OEDF is cyclic, hence  $\angle OEF = \angle ODF $ (angle subtended by the arc OF at the circumference).

Similarly $ \angle OFE = \angle ODE $.

But $ \angle OEF = \angle OFE $ (see hint 4).

Hence $ \angle ODE  = \angle ODF $. The proof is complete,

Reference: