Can You Solve This Cambridge Exam Question? | Number Theory | Infinite Descent

This problem comes from STEP 1 of the University of Cambridge entrance exam. Can you solve it?

The STEP exam papers can be found here:

Explain why a must be divisible by 3, and show further that both b and c must also be
divisible by 3. Hence show that the only integer solution is a = b = c = 0.

Did you solve it?
Let me know in the comments how you solved this problem. I’d love to hear about it!

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[promaxsavior]

The exposure to great questions is great sir 👌👌👍👍👏👏

[domc]

Surprisingly simple for step, thouggh first one can be a bit sneaky if you dont mention the fundamental theorem of arithmatic

[paschalisk]

Not sure about the second part of the solution. You show that all (a/3^n, b/3^n, c/3^n) will be solutions -- but this does not lead to any contradiction. The fact that we are looking for integers just means that only a few of those solutions (namely, the integer ones + zero) will be of interest. Am I missing something? In any case, a^3 can be written as 3^(3n1)p1, b^3 can be written as 3^(3n2)p2 and c^3 as 3^(3n3)p3, where p1, p2, p3 are not divisible by 3; then, the LHS is a multiple of the max of 3^(3n1) or 3^(3n2 + 1); while the RHS is a multiple of 3^(3n3+2) -- we have an imbalance of threes on one side, and, therefore, there exists no non-zero solution.

[onlypuremathematics]

Sir, My another question is can we get any parametric solution for an equation X^3 + 3Y^3 = Z^2 ? I found (X, Y, Z) =(1, 2, 5)

[VerSalieri]

I do not agree with your solution of the second part of the problem, how can you “choose” a to be 0 (I get that it has to be zero, but the justification for your statement seems shaky. .. Couldn’t we just use the infinite descent method? or argue by contradiction, taking a “minimal” solution where gcd(a, b) and 3 are coprime for example...