May 2025
The pair \(m = 3\), \(e = 1\), satisfies \(m^2 - 8e^2 = 1\). We can use this solution to find the pairs \(m\) and \(e\) that satisfy \(m^2 - 8e^2 = 4\) as follows: \[4 = 4(3^2 - 8(1^2)) = (2\cdot 3)^2 - 8(2\cdot 1)^2 = 6^2 - 8(2)^2.\] Similarly we have \[9 = 9(3^2 - 8(1^2)) = (3\cdot 3)^2 - 8(3\cdot 1)^2 = 9^2 - 8(3)^2.\] So, the pairs \((m,e) = (6,2)\) and \((m,e) = (9,3)\) satisfy \(m^2 - 8e^2 = 4\) and \(m^2 - 8e^2 = 9\) respectively.
The pair \(m = 9\) and \(e = 2\) satisfies \(m^2 - 8e^2 = 49\). There are lots of other pairs too. Here are a few others: \((m,e) = (11,3), (43, 15), (57, 20), (249,88)\).
It is tempting to just imitate what we did in Question 1. However, that yields the pair \((m,e) = (21,7)\), and \(21\) and \(7\) are not coprime!
We have \[\begin{align*} N((a + b\sqrt8)(c + d \sqrt 8)) &= N(ac + 8bd + (ad + bc)\sqrt 8) \\ &= (ac + 8bd + (ad + bc)\sqrt 8)(ac + 8bd - (ad + bc)\sqrt 8) \\ &= (ac + 8bd)^2 - 8(ad + bc)^2 \\ &= a^2c^2 + 16abcd + 64b^2d^2 -8a^2d^2 - 16abcd - 8b^2c^2 \\ &= (a^2 - 8b^2)(c^2 - 8d^2) \\ &= (a + b\sqrt 8)(a - b \sqrt 8)(c + d \sqrt 8)(c - d\sqrt 8) \\ &= N(a + b\sqrt 8)N(c + d\sqrt 8). \end{align*}\]
There are two important observations to make here. First, \(391 = 17 \times 23\). Second, is that \(N(a + b\sqrt8) = a^2 - 8b^2\). So, finding integers \(a,b\) satisfying \(a^2 - 8b^2 = d\) is equivalent to finding integers \(a,b\) satisfying \(N(a + b\sqrt 8) = d\).
From the information given in the problem statement, we have \(N(19 + 3\sqrt 8) = 17^2\) and \(N(27 + 5\sqrt 8) = 23^2\). Therefore, applying Question 3 we have \[\begin{align*} 391^2 = 17^2 \cdot 23^2 &= N(19 + 3\sqrt 8)N(27 + 5\sqrt 8) \\ &= N\big((19 + 3\sqrt 8)(27 + 5 \sqrt 8)\big) \\ &= N(633 + 176 \sqrt 8).\end{align*}\] Therefore \(633^2 - 8(176^2) = 391^2\). It remains to check that \(633\) and \(176\) are coprime.
The positive divisors of \(633\) are \(1, 3, 211, 633\). Since \(3, 211\), and \(633\) are not divisors of \(176\), we can conclude that \(1\) is the only positive common divisor of \(633\) and \(176\).
From Question 2, we know \(9^2 - 8\cdot 2^2 = 7^2\). Notice that \(N(9 + 2\sqrt 8) = 9^2 - 8\cdot 2^2 = 7\). So, with the result from Question 3 at our disposal, we can repeatedly square \(9 + 2\sqrt 8\) to get elements \(a_n\) and \(b_n\) of our sequence.
To that end, define \(a_1 = 9\) and \(b_1 = 2\). We then recursively define \(a_n\) and \(b_n\) by \[a_n + b_n \sqrt 8 = (a_{n-1} + b_{n-1}\sqrt 8)^2 = (a_{n-1}^2 + 8b_{n-1}^2) + 2a_{n-1}b_{n-1}\sqrt 8.\] Therefore \(a_n = a_{n-1}^2 + 8b_{n-1}^2\) and \(b_n = 2a_{n-1}b_{n-1}\). To see \(a_n^2 - 8b_n^2 = 7^{2^n}\) we have \[a_n^2 - 8b_n^2 = N(a_n + b_n\sqrt 8) = N((a_1 + b_1\sqrt 8)^{2^{n-1}}) = (N(a_1 + b_1\sqrt 8))^{2^{n-1}} = 7^{2^n}\] where the third equality is obtained by applying Question 3 repeatedly.
It remains to show that for every \(n\), \(a_n\) and \(b_n\) are coprime. To this end, first note that \(a_1\) and \(b_1\) are coprime (since \(a_1 = 9\) and \(b_1 = 2\)).
Next we will prove that if \(a_n\) and \(b_n\) share a prime factor \(p\) (which is equivalent to the statement that \(a_n\) and \(b_n\) are not coprime), then \(a_{n-1}\) and \(b_{n-1}\) must also share a prime factor \(p\). Once we have proved this, we can repeatedly apply it to show that if there is some \(n\) for which \(a_n\) and \(b_n\) share a prime factor, then \(a_1\) and \(b_1\) must share the same prime factor, a contradiction!
So, assume \(p\) is a prime that divides both \(a_n\) and \(b_n\). Since \(a_n^2 - 8b_n^2 = 7^{2^n}\), which is odd, \(a_n\) and \(b_n\) cannot both be even. Therefore \(p \neq 2\).
To proceed, we will repeatedly exploit the following two properties of prime numbers, which we now state without proof:
If \(p\) is a prime and \(p\) divides \(ab\), then \(p\) divides \(a\) or \(p\) divides \(b\).
If \(p\) is a prime and \(p\) divides \(n^2\), then \(p\) divides \(n\).
The first property is orten called Euclid’s lemma. Note that the second property is a special case of the first.
With these facts in our back pocket, let’s return to the proof. Since \(p\) is an odd prime and \(p\) divides \(2a_{n-1}b_{n-1}\), we must have that \(p\) divides \(a_{n-1}\) or \(b_{n-1}\).
Since \(p\) divides \(a_n\), write \(pm = a_n\) for some integer \(m\). Suppose first that \(p\) divides \(a_{n-1}\), so \(pk = a_{n-1}\) for some integer \(k\). Then \(pm = p^2k^2 + 8b_{n-1}^2\), which rearranges to \(8b_{n-1} = p(m - pk^2)\). Therefore \(p\) divides \(8b_{n-1}^2\). Since \(p\) is an odd prime, we must have that \(p\) divides \(b_{n-1}\).
On the other hand, suppose that \(p\) divides \(b_{n-1}\) and write \(pk = b_{n-1}\). Then similar to the previous case we have \(pm = a_{n-1}^2 + 8p^2k^2\). Therefore \(p\) also divides \(a_{n-1}\).
We have proved what we set out to prove: If \(a_n\) and \(b_n\) share a prime divisor \(p\), then \(a_{n-1}\) and \(b_{n-1}\) share the prime divisor \(p\). Repeatedly applying this result, we get that if \(a_n\) and \(b_n\) share a prime divisor \(p\), then \(a_1 = 9\) and \(b_1 = 2\) share a prime divisor \(p\). However, since \(2\) and \(9\) share no prime divisors, \(a_n\) and \(b_n\) share no prime divisors for all \(n\), completing the proof.
If you are familiar with induction, you can formalise the above argument as an inductive argument.
To finish things off, let’s explicitly calculate the first few terms in our sequences. We have \[\begin{align*} (a_1,b_1) &= (9,2) \\ (a_2,b_2) &= (113,36)\\ (a_3,b_3) &= (23\,137,8\,136)\\ (a_4,b_4) &= (1\,064\,876\,737,376\,485\,264).\end{align*}\] Sure enough, it turns out that \[(1\,064\,876\,737)^2 - 8(376\,485\,264)^2 = 33\,232\,930\,569\,601 = 7^{16}.\] Cool.