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$$13^k=a^2+b^2,k\in \mathbb{N}$$ has only one solution $(a,b)$ where $a,b\in \mathbb{N^{+}},gcd(a,b)=1$.

For example, $$13=2^2+3^2$$ $$13^2=169=5^2+12^2$$

I think this is very interesting problem. Thank you everyone.

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But there is always more than one integer solution...$13 = 2^2 + 3^2 = (-2)^2 + 3^2 = 2^2 + (-3)^2 = (-2)^2 + (-3)^2$. – fretty Aug 1 '13 at 9:12
The statement is false, e.g. $$13^3 = 26^2 + 39^2 = 9^2 + 46^2$$ – achille hui Aug 1 '13 at 9:19
What is true is that it has only one solution with $0\lt a\lt b$ and $\gcd(a,b)=1$. – Gerry Myerson Aug 1 '13 at 9:31
The solutions can be obtained from looking at the real and imaginary part of $(2+3i)^k$. – Hagen von Eitzen Aug 1 '13 at 9:33
Also, for even $k$, Hagen's solution is different from $13^k=\left(13^{\frac{k}{2}}\right)^2+0^2$, which gives an infinite number of counterexamples. – Tomas Aug 1 '13 at 9:42

The number of solutions of $13^{k} = a^{2} + b^{2}$ is given by "number of divisors of $13^{k}$ of the form $4n + 1$" minus "number of divisors of $13^{k}$ of the form $4n + 3$" and then multiply by $4$. Clearly $13^{k}$ has only $k + 1$ divisors of the form $4n + 1$ and no divisors of the form $4n + 3$, it follows that there are $4(k + 1)$ solutions of the equation $13^{k} = a^{2} + b^{2}$. Ignoring signs this becomes $k + 1$. Note that we are ignoring signs but counting $(a, b)$ and $(b, a)$ as two solutions, so that commutation is not ignored. So for the given question, unique solution is not guaranteed, but at least one solution is.

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Proof for number of solutions formula can be found at… – Paramanand Singh Aug 1 '13 at 9:33
It is not the fact, that $13^k$ is of the form $4n+1$, that guarantes a solution, but rather that $13$ is of the form $4n+1$. Note, that $21$ is of that form, but not sum of two squares, because $3,7$ are not of that form. – Tomas Aug 1 '13 at 9:35
Tomas you are correct. But as I explain the formula for number of solutions, if we try the number 21, we see that it has two divisors of the form $4n + 1$ (namely 1, 21) and two divisors of the form $4n + 3$ (3, 7) so that total number of solutions is zero. I will update my answer accordingly. Thanks for pointing out the silly mistake. – Paramanand Singh Aug 1 '13 at 9:38
Indeed! This is a very nice observation, I haven't seen it before (+1) – Tomas Aug 1 '13 at 9:40
You might add, that "Number of $\equiv 1$-divisors minus number of $\equiv 3$-divisors" yields the number of solutions ignoring signs, but not ignoring commutation. – Tomas Aug 1 '13 at 9:48

Known facts:

  1. $13$ is a prime of the form $4k+1$.
  2. For every prime $p$ of the from $4k+1$, there are integers $1 \le x < y < p$ such that $p = x^2 + y^2$. In particular, $13 = 2^2 + 3^2 = (2+3i)(2-3i)$.
  3. $\mathbb{Z}[i] = \{\;x + i y : x, y \in \mathbb{Z}\;\}$, the set of Guassian integers is a unique factorization domain with units $\pm 1, \pm i$ and three type of primes:

    1. $1+i\quad$ ( $1-i$ is equivalent to $1+i$ through a unit ).
    2. $x+iy, x-iy\quad$ where $x^2 + y^2 = p$ is a prime of the form $4k+1$.
      The $x \pm iy$ are inequivalent over $\mathbb{Z}[i]$.
    3. $p$ a prime of the form $4k+3$.

Since $13 = (2+3i)(2-3i)$, in any representation of $13^k$ as a sum of squares:

$$13^k = a^2 + b^2 = (a+bi)(a-bi)\tag{*1}$$ the unique factorization property of $\mathbb{Z}[i]$ forces $a + ib$ to have the form:

$$a + bi = i^e (2 + 3i)^f (2-3i)^{k-f}\tag{*2}$$

where $0 \le e < 4$ and $0 \le f \le k$.

If $0 < f < k$, then the R.H.S of $(*2)$ contains a factor $(2+3i)(2-3i) = 13$. This will make $\gcd(a,b) > 1$ (or one of $a,b$ vanish). This means if one want a solution of $(*1)$ with $\gcd(a,b) = 1$, $f$ can only be $0$ or $k$.

Notice the $f = 0$ case can be obtained from a $f = k$ case by taking complex conjugation. Furthermore, multiplication of $i^e$ either changes the signs of $a,b$ or exchange among them.

As a result, up to flipping the signs and/or exchaning $a$ with $b$, the unique solution of $(*1)$ with $\gcd(a,b) = 1$ is given by (as pointed by other poster):

$$a + i b = (2+3i)^k$$

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