Lesson 4

The answer is negative. To prove this, suppose, for the sake of contradiction, that such a quadruple of numbers exists. Take the quadruple \((a,b,c,d)\) where \(a\) is minimal. It is clear that \(d\) must be even. Then \(c\) must also be even: if not, \(2c^{4}\) would not be divisible by four, but everything else would be. Similarly, it can be shown that \(b\) and \(a\) must be even (divisibility by eight and sixteen, respectively). We obtain that all numbers in the quadruple are even, but then they can all be divided by two, resulting in a solution where \(a\) is smaller.