MHB Can Every Natural Number be Expressed as a Sum of Square Roots?

[anemone]

Prove that for any natural number $a$, there is an integer $k$ such that $(\sqrt{1982}+1)^a=\sqrt{k}+\sqrt{k+1981^a}$

 

[Nono713]

Spoiler

We can actually solve this for $k$ directly:
$$(\sqrt{1982}+1)^a=\sqrt{k}+\sqrt{k+1981^a}$$
$$(\sqrt{1982}+1)^a - \sqrt{k}=\sqrt{k+1981^a}$$
$$\left ((\sqrt{1982}+1)^a - \sqrt{k} \right )^2=k+1981^a$$
$$(\sqrt{1982}+1)^{2a} - 2 \sqrt{k} (\sqrt{1982}+1)^a + k=k+1981^a$$
$$(\sqrt{1982}+1)^{2a} - 2 \sqrt{k} (\sqrt{1982}+1)^a = 1981^a$$
$$2 \sqrt{k} (\sqrt{1982}+1)^a = (\sqrt{1982}+1)^{2a} - 1981^a$$
$$\sqrt{k} = \frac{(\sqrt{1982}+1)^{2a} - 1981^a}{2 (\sqrt{1982}+1)^a}$$
Now just note that:
$$\left (\sqrt{1982} - 1 \right ) \left (\sqrt{1982} + 1 \right ) = 1981$$
And so we get:
$$\sqrt{k} = \frac{(\sqrt{1982}+1)^{2a} - \left (\sqrt{1982} - 1 \right )^a \left (\sqrt{1982} + 1 \right )^a}{2 (\sqrt{1982}+1)^a}$$
We can now divide through to get:
$$\sqrt{k} = \frac{1}{2} \left [ (\sqrt{1982}+1)^a - \left (\sqrt{1982} - 1 \right )^a \right ]$$
And finally:
$$k = \frac{1}{4} \left [ (\sqrt{1982}+1)^a - \left (\sqrt{1982} - 1 \right )^a \right ]^2$$
Expanding:
$$k = \frac{1}{4} \left [ (\sqrt{1982}+1)^{2a} + \left (\sqrt{1982} - 1 \right )^{2a} - 2 \cdot 1981^a \right ]$$
Finally we use the binomial theorem to expand the powers:
$$\left ( \sqrt{1982} + 1 \right )^{2a} + \left (\sqrt{1982} - 1 \right )^{2a} = \sum_{k = 0}^{2a} \binom{2a}{k} \left [ \sqrt{1982}^{2a - k} + \sqrt{1982}^{2a - k} (-1)^k \right ]$$
Now for odd $k$ the $k$th term vanishes, so consider only even $k$, say $k = 2m$:
$$\left ( \sqrt{1982} + 1 \right )^{2a} + \left (\sqrt{1982} - 1 \right )^{2a} = \sum_{m = 0}^{a} \binom{2a}{2m} \sqrt{1982}^{2a - 2m} + \sqrt{1982}^{2a - 2m} = \sum_{m = 0}^{a} \binom{2a}{2m} 2 \cdot 1982^{a - m}$$
So that the sum is clearly an integer. Note that 1982 is even, so that all the terms in the series except the last (which is always just $(2a, 2a) 2 \cdot 1982^0 = 2$) are divisible by 4, so that the sum as a whole is divisible by 2 but not by 4. On the other hand, $2 \cdot 1981^a$ is also divisible by 2 but not by 4, and so:
$$(\sqrt{1982}+1)^{2a} + \left (\sqrt{1982} - 1 \right )^{2a} - 2 \cdot 1981^a ~ ~ ~ \text{is divisible by 4}$$
We conclude that:
$$k = \frac{1}{4} \left [ (\sqrt{1982}+1)^{2a} + \left (\sqrt{1982} - 1 \right )^{2a} - 2 \cdot 1981^a \right ] = \frac{1}{4} \left [ (\sqrt{1982}+1)^a - \left (\sqrt{1982} - 1 \right )^a \right ]^2$$
is in fact an integer and a solution to the original equation as derived earlier. $\blacksquare$

First few solutions $(a, k)$ are given by $(0, 0), (1, 1), (2, 7928), (3, 35366809), (4, 124700748768), \cdots$

 

[anemone]

Hi Bacterius, sorry for the late reply, and thanks for the great solution!:)

I also want to share the solution (of other) here:

Spoiler

Let $$X=\sum_{j=o\,\,j\,\,\text{even}}^{a} {a\choose j}(\sqrt{1982})^j$$, $$Y=\sum_{j=o\,\,j\,\,\text{odd}}^{a} {a\choose j}(\sqrt{1982})^j$$

Thus $X$ is the sum of the even-numbered terms in the expansion of $(\sqrt{1982}+1)^a$ and $Y$ is the sum of the odd-numbered terms in that expansion.

Note also that

$$(\sqrt{1982}-1)^a=\sum_{j=o}^{a} {a\choose j}(\sqrt{1982})^{a-j}(-1)^j=\begin{cases}Y-X \,\,\,\text{if}\,\,a\,\,\text{is odd}, & \\[3pt] X-Y \,\,\,\text{if}\,\,a\,\,\text{is even} \\ \end{cases}$$

Case 1: $a$ is odd. Let $k=X^2$

Then we have

$\begin{align*}\sqrt{k+1981^a}+\sqrt{k}&=\sqrt{X^2+(\sqrt{1982}-1)^a(\sqrt{1982}+1)^a}+X\\&=\sqrt{X^2+(Y-X)(X+Y)}+X\\&=\sqrt{X^2+Y^2-X^2}+X\\&=Y+X\\&=(\sqrt{1981}+1)^a \end{align*}$

Case 2: $a$ is even. Let $k=Y^2$

Then we have

$\begin{align*}\sqrt{k+1981^a}+\sqrt{k}&=\sqrt{Y^2+(X-Y)(X+Y)}+\sqrt{Y^2}\\&=\sqrt{X^2}+\sqrt{Y^2}\\&=(\sqrt{1981}+1)^a \end{align*}$

It's evident that $X^2$ is an integer, and $Y^2$ is an integer since $Y$ has the form $\sqrt{1982}M$, for some integer $M$, and this completes the proof.