The article didn't explain the abc conjecture in much detail ("The original proof is of the long-standing ABC conjecture that explores the deep nature of number

The article didn't explain the abc conjecture in much detail ("The original proof is of the long-standing ABC conjecture that explores the deep nature of numbers, centred on the simple equation a + b = c."). That's unfortunate, because even though the proof is inscrutable, the conjecture itself isn't very complicated: If three numbers a, b, and c are coprime and two numbers a and b have large prime factors, then their sum c generally does not. To make that a little more formal, it means that the radical (product of distinct prime factors) of their product abc is usually less than the sum c, up to some exponent 1+eps: rad(abc)^(1+eps) < c ... for some eps. That eps is important! There are infinitely many numbers a, b, c where this isn't true for some eps (and there are ways to construct those), but the conjecture states that for any given eps, there are finitely many. To find maximal counterexamples (which, because mathematics consistently uses intuitive terminology, are said to be "high-quality") you typically look for numbers that are smooth (small prime factors) but not powersmooth (larger prime-factors-raised-to-a-power). Typically you'll also fix a to be really small, to limit the search space somewhat. [MM14] describes other methods for finding triples. Here's a simple example of a high-quality triple: a = 1 b = 2*3^7 = 4374 c = 5^4*7 = 4375 Because all factors are unique, rad(abc) = rad(a) rad(b) rad(c). c > (rad(abc))^(1+eps) c > (rad(a) * rad(b) * rad(c))^(1+eps) 5^4*7 > (rad(1) * rad(2*3^7) * rad(5^4*7))^(1+eps) 5^4*7 > (1 * 2 * 3 * 5 * 7)^(1+eps) 4375 > (210)^(1+eps) This is a great example, because the exponent (sometimes called q) is about 1.57, which is extraordinarily high. In the first billion c's, there are only 34 with q > 1.4. This is the fifth best such triple we know of. [triples] Consider a trivial case where all three numbers are prime; a = 2, b = 3, c = 5. Primes are their own radical, so: 5 > (2 + 3)^1+eps This is the edge case. [triples]: https://www.math.leidenuniv.nl/~desmit/abc/index.php?set=2 https://www.math.leidenuniv.nl/~desmit/abc/index.php?set=2 [MM14]: https://arxiv.org/pdf/1409.2974.pdf https://arxiv.org/pdf/1409.2974.pdf