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One of my favorite math problems that is easy to solve with just algebra:
Prove that 8 is the only perfect cube to follow a prime number.
If you don't know what a perfect cube is, that is simple, it is any integer raised to the power of 3. Since \(8 = 2^3\) it is a perfect cube, and it follows the number 7, which is prime. 8 is the only number that fits those conditions... prove it.
NOTE: I will give the answer as a reply. If anyone else wants to provide an answer please make sure you use a content warning.
Answer
Assume the opposite: some cube a³ = p + 1 for some natural number a ≠ 2 and prime p.
The difference of cubes formula shows that:
p = a³ - 1 = (a² + a + 1)(a - 1)
Both terms (a² + a + 1) and (a - 1) are integers. Since p is prime, it follows that exactly one of the two must be equal to unity.
The first possibility can be ruled out because a² + a + 1 = 1 has no solution in the natural numbers.
The second is impossible because a - 1 = 1 contradicts the assumption a ≠ 2.
Further discussion
The case where a is an integer but not a natural number can be excluded because primes are a subset of the natural numbers. If a < 1, then p = a³ - 1 < 0 which contradicts p's primality.
