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Explore the Back-StoryRMO 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.

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

- Number Theory:
- Understand the notion of modular inverse.
- Bezout's Theorem.

Also see

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.

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 $$ a = 3^{20k}, b = 3^{15k}, c = 3^{12k} $$

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: $$ 3^{300} + 3^{300} + 3^{300} = (3^{100})^3 + (3^{75})^4 + (3^{60})^5 = 3^{301} = (3^{43})^7 $$

- These ideas are usually discussed in the
**Number Theory I**module of Cheenta Math Olympiad Program. **Elementary Number Theory by David Burton i**s a good reference for some these ideas.

Also see

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.

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

- Number Theory:
- Understand the notion of modular inverse.
- Bezout's Theorem.

Also see

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.

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 $$ a = 3^{20k}, b = 3^{15k}, c = 3^{12k} $$

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: $$ 3^{300} + 3^{300} + 3^{300} = (3^{100})^3 + (3^{75})^4 + (3^{60})^5 = 3^{301} = (3^{43})^7 $$

- These ideas are usually discussed in the
**Number Theory I**module of Cheenta Math Olympiad Program. **Elementary Number Theory by David Burton i**s a good reference for some these ideas.

Also see

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