2 problems found
Find the integer, \(n\), that satisfies \(n^2 < 33\,127< (n+1)^2\). Find also a small integer \(m\) such that \((n+m)^2-33\,127\) is a perfect square. Hence express \(33\,127\) in the form \(pq\), where \(p\) and \(q\) are integers greater than \(1\). By considering the possible factorisations of \(33\, 127\), show that there are exactly two values of \(m\) for which \((n+m)^2 -33\,127\) is a perfect square, and find the other value.
Solution: \begin{align*} 180^2 &= 32400 \\ 181^2 &= 32761 \\ 182^2 &= 33124 \\ 183^2 &= 33489 \\ 184^2 &= 33856 \end{align*} Therefore \(182^2 < 33\,127 < (182+1)^2\). and \((182+2)^2 - 33\,127 = 729 = 27^2\). Therefore \(33\,127 = 184^2 - 27^2 = 211 \times 157\). (Note both of these numbers are prime). Suppose \((n+m)^2 - 33\,127 = k^2\) then \(33\,127 = (n+m)^2-k^2 = (n+m-k)(n+m+k)\). Since there are only two factorisations of \(33\,127\) into positive integer factors with one factor larger than the other, the other factorisation must be: \(n+m+k = 33\,127, n+m-k = 1 \Rightarrow k = \frac{33\, 126}{2} = 16563\), ie \(16564^2 - 33\,127 = 16563^2\)
Find the integers \(k\) satisfying the inequality \(k\leqslant2(k-2).\) Given that \(N\) is a strictly positive integer consider the problem of finding strictly positive integers whose sum is \(N\) and whose product is as large as possible. Call this largest possible product \(P(N).\) Show that \(P(5)=2\times3, P(6)=3^{2}, P(7)=2^{2}\times3, P(8)=2\times3^{2}\) and \(P(9)=3^{3}.\) Find \(P(1000)\) explaining your reasoning carefully.
Solution: \begin{align*} && k &\leq 2(k-2) \\ \Rightarrow && 4 &\leq k \end{align*} Lemma: Suppose we construct \(N \neq \) (optimally) as a sum out of \(a_1 + \cdots +a_k\), then \(a_i \in \{2, 3\}\). Proof: Suppose not, suppose some \(a_i > 3\). Then from our earlier inequality, the sum \(a_1 + \cdots +a_{i-1} + 2 + (a_i - 2) + \cdots \) has the same sum, but a larger product. Therefore \(a_i \leq 3\). Suppose also some \(a_i = 1\), then we could replace \(a_1\) with \(a_1+1\) and remove \(a_i\), leaving us again with the same sum but larger product. (Assuming \(N \neq 1\)) \(5 = 2+3\) is the only way to write \(5\) as a sum of \(2\)s and \(3\)s, therefore \(P(5) = 2\times 3\) \(6 = 2 + 2 + 2 = 3 + 3\) and we can immediately see that \(2^3 = 8 < 3^2 = 9\), so \(P(6) = 3^2\) and whenever we have three \(2\)s we should replace them with two \(3\)s. So \(7 = 2 + 2 + 3 \Rightarrow P(7) = 2^2 \times 3\) \(8 = 3 + 3 + 2 \Rightarrow P(8) = 2\times 3^2\) \(9 = 3 + 3 + 3 \Rightarrow P(9) = 3^3\) Suppose \(1000 = 2n + 3m\), considered \(\pmod{3}\) we can see that \(n \equiv 2 \pmod{3}\) therefore we should have \(1000 = 2 + 2 + \underbrace{3 + \cdots + 3}_{332\text{ }3\text{s}}\) and so \(P(1000) = 2^2 \times 3^{332}\)