Alan

You can factor it as:

$$(x-\frac{1}{2})(x-4m+\frac{1}{2})=0$$

so  

$$\\ x=\frac{1}{2}\\ or\\x=4m-\frac{1}{2}$$

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This will be true if x⁴ + 2 = x² + 4  or  x⁴ - x² - 2 = 0   or  (x² + 1)(x² - 2) = 0

The only option that gives real solutions for x is  x² = 2 or x = ±√2

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Here's a proof that only the three digits 1, 2 and 3 are involved:

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This is more engineering than physics, and I'm not sure I've made the simplifying assumptions the questioner was looking for!  Perhaps he/she will tell us.

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Does this help:

I've mistakenly used 200m for L instead of 1200m above.  With 1200m the flowrate becomes F ≈ 3050 kg/s.

Do you mean the 3's to go on forever?  If so, then try 604/300

$${\frac{{\mathtt{604}}}{{\mathtt{300}}}} = {\frac{{\mathtt{151}}}{{\mathtt{75}}}} = {\mathtt{2.013\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

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$${{\mathtt{\,-\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{5}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{32}}}{{\mathtt{243}}}} = -{\mathtt{0.131\: \!687\: \!242\: \!798\: \!353\: \!9}}$$

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This reduces to 600pi = (1/2)1800*theta or 600pi = 900*theta or theta = 2pi/3

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2^4 is 16, so this reduces to 2^(6x)/2^(x^2)=1  or 2^(6x) = 2^(x^2)  so that 6x = x^2  so x = 0 or x = 6

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