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USDescartes posed the following problem [1]:

> Let $x$ be any number that makes $4^{3141} + 4^x + 4^{1618}$ a perfect
square. Find the sum $S$ of all such values of $x$.

#### Approach

We will rewrite USDescartes' expression so that the base of each power is 2. We will then apply a theorem of Szalay to find all the values of $x$ for which the expression is a perfect square.

#### Background

By a theorem of Szalay [2], we have:

> If the positive integers $n$, $m$ and $x$  with $n$ > $m$ satisfy
> $$2^n+2^m + 1 = x^2$$
>
> then
>
> (i) $(n,m,x) \in \{(2t, t + 1, 2^t + 1) | t \in \mathbb{N}, t \ge 1\}$ or
>
> (ii) $(n,m,x) \in \{(5,4,7), (9,4,23)\}$

#### Changing the base of each term from 4 to 2

Let $y = 2x$. Then the following expressions are equal:

$$\begin{align}
&\hspace{1.5em}4^{3141} + 4^x + 4^{1618} \\
&=2^{6282} + 2^y + 2^{3236}
\end{align}
$$

If $y$ is not a nonnegative integer, (2) will not be a nonnegative integer, and hence (2) cannot be a perfect square. We therefore conclude that $y$ must be a nonnegative integer. Additionally, the large exponents rule out case (ii) of Szalay's theorem, so we will proceed by cases using case (i).

#### Case 1: $3236 > y \ge 0$

Here, $2^y$ is the least term of (2), so it can be factored out of all three terms of (2), yielding a pair of integer factors:

$$\begin{align}2^y(2^{6282-y} + 2^{3236-y} + 1)\end{align}$$

Since $2^y$ is a power of 2 and the factor in parentheses is odd, both factors are relatively prime. Together with the fact that (3) is a perfect square, this implies that the prime factors of both factors must appear to an even power in the prime factorization of (3). This, in turn, implies that if (3) is a perfect square, then the result of dividing (3) by $2^y$ must also be a perfect square.

Hence we want to find the values of $y$ that make the following expression a perfect square:

$$\begin{align}2^{6282-y} + 2^{3236-y} + 1\end{align}$$

We can apply case (i) of Szalay's theorem to (4) to obtain the following pair of linear equations:

$$\begin{align*}6282-y &= 2t\\
3236-y &= t + 1
\end{align*}$$

Each solution to this pair of equations determines a value for $y$ that makes (4), and, by extension, (1), a perfect square. Solving, we find $t = 3047$ and $y = 2x = 188$. Hence our first value for $x$ that makes (1) a perfect square is $94$.

#### Case 2: $6282 > y \ge 3236$

Recall expression (2):
$$2^{6282} + 2^y + 2^{3236}$$

In this case, $2^{6282}$ is the largest term of (2) and $2^{3236}$ is the smallest. We can hence factor $2^{3236}$ out of all three terms of (2), yielding:

$$\begin{align}2^{3236}(2^{3046} + 2^{y-3236} + 1)\end{align}$$

By reasoning similar to that used in case 1, we can drop the $2^{3236}$ factor and what remains will still be a perfect square just when (5) is:

$$\begin{align}2^{3046} + 2^{y-3236} + 1\end{align}$$

We can then apply case (i) of Szalay's theorem to (6) to obtain the following pair of linear equations:

$$\begin{align*}
3046 &= 2t\\
y-3236 &= t + 1
\end{align*}$$

Each solution to this pair of equations determines a value for $y$ that makes (6), and, by extension, (1), a perfect square. Solving, we find that $t = 1523$ and $y = 2x = 4760$. Hence our second value for $x$ that makes (1) a perfect square is $2380$.

#### Case 3: $y \ge 6282$

As in the previous case, $2^{3236}$ is still the smallest term in (2).
Hence the properties of expression (6) from case 2 apply equally in this case; however, the $2^{y-3236}$ term is now the largest (or tied for largest).

Rearranging the terms in (6) for clarity to put the largest term first, we obtain:

$$\begin{align}2^{y-3236} + 2^{3046} + 1\end{align}$$

We now apply case (i) of Szalay's theorem to (7) to obtain the following pair of linear equations:

$$\begin{align*}
y-3236 &= 2t\\
3046 &= t + 1
\end{align*}$$

Each solution to this pair of equations determines a value for $y$ that makes (7), and, by extension, (1), a perfect square. Solving, we find that $t = 3045$ and $y = 2x = 9326$. Hence $4663$ is the third and final value for $x$ that makes (1) a perfect square.

#### Conclusion

We have found that :

$$\begin{align*}
4^{3141} + 4^{94} + 4^{1618} &= (2^{94}(2^{3047} + 1))^2\\
4^{3141} + 4^{2380} + 4^{1618} &= (2^{1618}(2^{1523} + 1))^2\\
4^{3141} + 4^{4663} + 4^{1618} &= (2^{1618}(2^{3045} + 1))^2
\end{align*}$$

The case analysis was exhaustive, and so, by Szalay's theorem, there can be no other values for $x$ that make (1) a perfect square. Hence there are three numbers $x$ that make $4^{3141} + 4^x + 4^{1618}$ a perfect square: 94, 2380, and 4663. To complete the solution of the original problem, we note that the sum of these three is 7137. $\square$

[1] 
USDescartes [@nandor117]. "#POTD #Math #PerfectSquares Problem of the Day!" *Twitter* (2023-03-15) twitter.com/nandor117/status/1636051972825530377

[2] Lászó Szalay, The equations $2^N \pm 2^M \pm 2^L = z^2$, *Indagationes Mathematicae*,
Volume 13, Issue 1 (2002), Pages 131-142, ISSN 0019-3577,
doi.org/10.1016/S0019-3577(02)90011-X
(sciencedirect.com/science/article/pii/S001935770290011X)

Thu, 16 Mar 2023 03:50 GMT