Exploring a Vietnamese Number Theory Challenge
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Chapter 1: Introduction to the Problem
In this article, we delve into an intriguing number theory challenge originating from Vietnam! While only basic mathematical principles are necessary, the solution may not be immediately apparent. Are you able to crack it?
To begin with, we observe that the right-hand side of the equation is even. This indicates that both x and y must share the same parity; if x is even, then y must also be even, and if x is odd, y must follow suit. If they didn't share parity, the result of x³ - y³ would be odd, leading to a contradiction.
Now, let's factor the left-hand side. A well-known factorization can be expressed as follows:
If you're skeptical about this (even after expanding the right-hand side), try to convince yourself that this holds for any general exponent by treating y as a constant and executing long division by x - y.
Since both x and y possess the same parity, x - y is also even. Therefore, we can divide everything by two, leading us to:
The next step might not be immediately clear. For now, let's assume that xy > -4. This will become clearer shortly.
We deduce that x² + xy + y² divides xy + 4, and since xy + 4 is positive, it follows that x² + xy + y² ≤ xy + 4, which implies that x² + y² ≤ 4. The only pairs (x, y) satisfying this are (0, -2) and (2, 0).
However, we made an assumption that needs verification. Let's explore the validity of this assumption. Notice that if xy = 0, we obtain the solutions we’ve identified. Now, let’s consider the case where xy is non-zero.
Assuming x > 0 > y, we find that -y³ and -2xy are both positive. Consequently, we are constrained to a very limited region for the original equation:
If either -y³ or -x³ is greater than or equal to 2, they exceed 8, leading to no solutions in the case where x > 0 > y.
Now, consider the scenario where y > 0 > x, and let z = -x > 0 for clarity. Our equation can be reformulated as:
Without loss of generality, let's assume that z ≤ y and y ≥ 2. Therefore, we have z ≥ 2 and z² ≤ yz. By multiplying these inequalities, we conclude that z³ ≤ 2yz. Substituting this back into our equation yields:
y³ + z³ + 8 ≤ z³, which implies y³ + 8 ≤ 0. Since y is positive, this scenario is impossible.
What if y = 1? Then we have z³ + 8 = 2z. If this polynomial has a rational root, it must divide 8, but a quick check reveals that it does not have any rational roots. Thus, we conclude that if xy < 0, no solutions exist.
Ultimately, the only solutions we derive are:
Hopefully, this problem proved to be a stimulating challenge! If you enjoyed this article, feel free to support my work.