# Find this this diophantine equation the number

Let $a,b$ be positive integer numbers. Find the number of pairs $(a,b)$ satisfying $$\dfrac{ab}{1998}=\sqrt{a^2+b^2}+a+b.$$

• I got a quick answer: $(1998.3,1998.4)$. – GAVD Aug 13 '15 at 10:12
• @MichaelGaluza: Sorry, I mean $(1998 \times 3, 1998 \times 4)$. – GAVD Aug 13 '15 at 10:41

At first, $\sqrt{a^2+b^2}$ must be integer. So, $a$ and $b$ come from Pythagorean triples and $$a = d(n^2 - m^2),\quad b=d\cdot2nm$$ for some integers $d$, $n$, $m$ ($n > m$). Hence $\sqrt{a^2+b^2}=d(n^2 + m^2)$, and then $$d\frac{(n-m)(n+m)\cdot 2nm}{1998} = n^2 + m^2 + n^2 - m^2 + 2nm=2n(n+m),$$ or $$dm(n-m)=1998\Longrightarrow n = m + \frac{1998}{md}.$$
Number of $(n,m)$-pairs is $$\sum_{d\mid 1998} \sigma_0\Big(\frac{1998}{d}\Big)=\\ = 16+ 8+ 12+ 6+ 8+ 4+ 4+ 8+ 2+ 4+ 6+ 3+ 4+ 2+ 2+ 1=90,$$ where $\sigma_0(n)$ is the number of the divisors of $n$. But $(a, b)$ can repeats; I found $42$ (really $42$!) pairs ($(a,b)$ and $(b,a)$ are the same pairs, so there is $84$ solutions): $$(3997, 7988004), (3999, 2665332), (4000, 1999998), (4005, 891108), (4008, 669330), (4023, 299700), (4032, 225774), (4033, 219780), (4077, 102564), (4104, 77922), (4107, 75924), (4144, 57942), (4239, 36852), (4320, 28638), (4329, 27972), (4440, 21978), (4725, 14948), (4968, 12210), (4995, 11988), (5328, 9990), (5365, 9828), (6912, 6734), (6993, 6660), (7992, 5994), (8103, 5940), (9472, 5454), (12987, 4884), (15984, 4662), (16317, 4644), (20424, 4482), (30969, 4292), (39960, 4218), (40959, 4212), (53280, 4158), (111888, 4070), (114885, 4068), (151848, 4050), (336663, 4020), (447552, 4014), (1001997, 4004), (1334664, 4002), (3996000, 3998)$$
• Thank you,+1 and can you find the last result? I use your last for to find the numeber is $42?$,I'm wrong? – user246688 Aug 13 '15 at 10:47
• your $90$ is contain some same pairs – user246688 Aug 13 '15 at 10:59