Let $a = \sqrt {75025} + \sqrt {121393} + \sqrt {196418} + \sqrt{317811}$ and $b = \sqrt {514229} + \sqrt {832040}$. By using SageMath we can see that $$ a - b \approx 2.95301560981898 \cdot 10^{-9} $$ That means almost nine digits of accuracy!

To investigate any particular reason why these surprising digits of accuracy come I considered the function $$ f(x)= \sqrt{x +317811} + \sqrt{x + 196418} + \sqrt{x + 121393} + \sqrt{x + 75025} -\sqrt{x + 832040} - \sqrt{x + 514229}$$

The Taylor series of $f$ around $x =0$ with approximate coefficients looks like $$ f(x) = f(0) + 0.00403020948350145x -2.13362649294736 \cdot 10^{-8}\frac12 x^2 + O(x^3) $$

If $\alpha$ is a root of the equation $f(x) = 0$ where $\alpha $ is very close to $0$ (definitely there is a root between $-1$ and $0$) then $$0 = f(\alpha) = f(0) + 0.00403020948350145 \alpha -2.13362649294736 \cdot 10^{-8}\frac12 \alpha^2 +\text{higher error terms} $$

Of course, we can use computer programs to find a bound on $\alpha$ but the whole process is not mathematically elegant.

Certainly, $\alpha$ is a root of some polynomial of higher degree, so it may be difficult to find an expression in terms of radicals for $\alpha$ or may not be possible at all.

Are there any other reasons for this level of accuracy?

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    $\begingroup$ How did you come across these numbers? $\endgroup$
    – lhf
    Jun 27, 2021 at 10:41
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    $\begingroup$ Could you add some context? How did you arrive at these numbers? $\endgroup$ Jun 27, 2021 at 10:42
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    $\begingroup$ @lhf My friend gave me (as a lockdown challenge ) who always uses a computer programs to generate this kind of expression. $\endgroup$ Jun 27, 2021 at 11:02
  • $\begingroup$ Uh, aren't those radicands Fibonacci numbers? $\endgroup$ Jul 12, 2021 at 20:30
  • $\begingroup$ Let's look at $75025$. Multiply this by $\sqrt5$ and take its natural log ~ 12.0303. The natural log of 1.61803... is ~ 0.481212. Divide the former by the latter and the result from my mobile device calculator is $25+10^{-10}$ ... aha! $\endgroup$ Jul 12, 2021 at 20:35

1 Answer 1


Let $x=75025$ and $y = 121393$. This approximate equality can be written as $$\sqrt x+\sqrt y+\sqrt{x+y} +\sqrt{x+2y}\approx\sqrt{2x+3y}+\sqrt{3x+5y}$$ where $y\approx x*\varphi$ and $\varphi$ is the golden ratio.

Factoring out $x$ from both sides, we get the approximate inequality as $$1+\sqrt\varphi+\sqrt{1+\varphi}+\sqrt{1+2\varphi}\approx\sqrt{2+3\varphi}+\sqrt{3+5\varphi}$$ Now consider the LHS and RHS separately (using the fact that $\varphi^2=\varphi+1$), \begin{align} LHS &= 1+\sqrt\varphi+\sqrt{1+\varphi}+\sqrt{\varphi+\varphi^2}\\ &= (1+\sqrt\varphi)(1+\sqrt{1+\varphi})\\ &=(1+\sqrt\varphi)(1+\varphi)\\ &=\varphi^2(1+\sqrt\varphi) \end{align} Also, \begin{align} RHS &= \sqrt{2\varphi^2+\varphi}+\sqrt{3\varphi^2+2\varphi}\\ &=\sqrt\varphi\sqrt{\varphi+\varphi^2}+\sqrt\varphi\sqrt{2\varphi^2+\varphi}\\ &=\varphi\sqrt{1+\varphi}+\varphi\sqrt{\varphi+\varphi^2}\\ &=\varphi^2(1+\sqrt\varphi) \end{align} So these expressions for golden ratio are exactly equal.

This is the reason we're getting approximate inequality in this case (because $y$ is not exactly equal to $x\varphi$, otherwise the given equality would also be exact).


I have written a function in python to find the difference between LHS and RHS of given equation for any given input $x$ (which is $75025$ in this case):

from math import sqrt

def find_diff(x):
    phi = (1+sqrt(5))/2
    y = round(x*phi)
    a = sqrt(x)+sqrt(y)+sqrt(x+y)+sqrt(x+2*y)
    b = sqrt(2*x+3*y)+sqrt(3*x+5*y)
    return abs(a-b)

Note that you may try further Fibonacci numbers and get better approximations (it is obvious to see what Fibonacci numbers are doing here if you know the continued fraction approximations for golden ratio).

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    $\begingroup$ (+1). But OP should have added some context, this would have had been a much simpler problem to solve, then. $\endgroup$ Jun 27, 2021 at 11:03
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    $\begingroup$ You are right @Ritam_Dasgupta, but there would be lesser fun to solve that simpler problem then :) $\endgroup$
    – Martund
    Jun 27, 2021 at 11:11
  • $\begingroup$ Use $\varphi ^{n}=F_{n}\varphi +F_{n-1}$ to simplify the derivation. $\endgroup$
    – lhf
    Jun 28, 2021 at 16:20
  • $\begingroup$ Actually the radicands are consecutive Fibonacci numbers. We should work this into the answer. $\endgroup$ Jul 12, 2021 at 20:39

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