The answer is $231361$ : $a = 14911,$ $b = 481.$ $15 \cdot 14911 + 16 \cdot 481 = 231361 = 481^2,$ $16 \cdot 14911 - 15 \cdot 481 = 231361 = 481^2.$ Now, let us prove it. We have : $E_1 : 15a + 16b = x^2,$ $E_2 : 16a - 15b = y^2.$ $E_1^2 + E_2^2 \implies 481(a^2+b^2) = x^4 + y^4.$ So $x^4 + y^4 = 0 \mod 13 \cdot 37.$ But, this implies $x^4 + y^4 = 0 \mod 13$ and $x^4 + y^4 = 0 \mod 37$ that is, since these are primes, and taking $x$ and $y$ to be non-zero in $\mathbb{Z}/13\mathbb{Z}$ and $\mathbb{Z}/37\mathbb{Z}$ : $\left(\frac xy \right)^4 = -1 \mod 13$ and $\left(\frac xy \right)^4 = -1 \mod 37.$ But, this is impossible, which implies that $x$ and $y$ are both divisible by $13$ and $37,$ hence $x = 481u$ and $u = 481v,$ hence $a^2 + b^2 = 481^3(u^2+v^2).$ The possible minimum for $\min(u, v)$ is $1.$ But, $1^2 + 31^2 = 2 \cdot 481 \implies 481^2 + (31 \cdot 481)^2 = 481^3(1^2 + 1^2).$ And, (today's a lucky day ), this necessary condition happens to be sufficient, giving the result announced in the very beginning of this post.

Peter wrote: The positive integers $ a$ and $ b$ are such that the numbers $ 15a + 16b$ and $ 16a - 15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares? We characterize the set \[ \mathcal{S} = \{\; (a,b)\;\vert\; 15a + 16b = x^{2},\; 16a - 15b = y^{2}\;\text{for some}\; x,y\in\mathbb{N }\;\}. \] Let $ (a,b)\in\mathcal{S}$. We can find positive integers $ x$ and $ y$ such that $ 15a + 16b = x^{2}$ and $ 16a - 15b = y^{2}$. After solving the system of equations $ 15a + 16b = x^{2}$, $ 16a - 15b = y^{2}$, we obtain \[ a = \frac { 15x^{2} + 16y^{2}}{15^{2} + 16^{2}} = \frac { 15x^{2} + 16y^{2}}{13\cdot 17},\; b = \frac {16x^{2} - 15y^{2}}{15^{2} + 16^{2}} = \frac {16x^{2} - 15y^{2}}{13\cdot 17}. \] Now, we have to work modulo $ 13$ and modulo $ 17$. Step 1. First, we work modulo $ 13$. We need to solve the system of congruences \[ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 13)\;\;\text{and}\;\; 16x^{2} - 15y^{2}\equiv 0\; (mod\; 13). \] The first equation reads $ 2x^{2} + 4y^{2}\equiv 0\; (mod\; 13)$ or $ x^{2}\equiv - 2y^{2}\; (mod\; 13)$. The second equation reads $ 3x^{2} - 2y^{2}\equiv 0\; (mod\; 13)$ or $ 3x^{2}\equiv 2y^{2}\; (mod\; 13)$. We apply Gauss Elimination. We compute $ 3 ( - 2y^{2})\equiv 2y^{2}\; (mod\; 13)$ or $ 8y^{2}\equiv 0\; (mod\; 13)$. This implies that $ y\equiv 0\; (mod\; 13)$. Also, we find that $ x^{2}\equiv - 2y^{2}\equiv 0\; (mod\; 13)$ so that $ x\equiv 0\; (mod\; 13)$. We conclude that $ x\equiv y\equiv 0\; (mod\; 13)$ is the unique solution of the system of congruences. \textsc{Step 2.} Second, we work modulo $ 17$. We need to solve the system of congruences \[ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 17)\;\;\text{and}\;\; 16x^{2} - 15y^{2}\equiv 0\; (mod\; 17). \] The first equation reads $ - 2x^{2} - y^{2}\equiv 0\; (mod\; 17)$ or $ 2x^{2}\equiv y^{2}\; (mod\; 17)$. The second equation reads $ - x^{2} + 2y^{2}\equiv 0\; (mod\; 17)$ or $ x^{2}\equiv 2y^{2}\; (mod\; 17)$. Hence, we compute $ 2 ( 2y^{2})\equiv y^{2}\; (mod\; 17)$ or $ 4y^{2}\equiv 0\; (mod\; 17)$. This implies that $ y\equiv 0\; (mod\; 17)$. Also, we find that $ x^{2}\equiv 2y^{2}\equiv 0\; (mod\; 17)$ so that $ x\equiv 0\; (mod\; 17)$. We conclude that $ x\equiv y\equiv 0\; (mod\; 17)$ is the unique solution of the system of congruences. Combining results from Step 1 and Step 2, we can write $ x = 13\cdot 17 u$ and $ y = 13\cdot 17 v$ for some positive integers $ u$ and $ v$. Now, observe that \[ 15a + 16b = x^{2},\; 16a - 15b = y^{2} \] becomes \[ a = \frac { 15x^{2} + 16y^{2}}{13\cdot 17},\; b = \frac {16x^{2} - 15y^{2}}{13\cdot 17} \] or \[ a = 13\cdot 17 (15u^{2} + 16v^{2}) ,\; b = 13\cdot 17 (16u^{2} - 15v^{2}). \] Hence, it follows that the set $ \mathcal{S}$ can be represented as \[ \mathcal{S} = \{\; (a,b)\;\vert\; a = 13\cdot 17 (15u^{2} + 16v^{2}) ,\; b = 13\cdot 17 (16u^{2} - 15v^{2}), u,v\in\mathbb{N }\;\}. \] We therefore conclude that the minimum value is $ \left( 13\cdot 17\right)^{2}$. Notes In the solution, we met the systems of congruences. We recall the congruences from Step 1: $ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 13)$, $ 16x^{2} - 15y^{2}\equiv 0\; (mod\; 13)$. There are many alternative ways to reach $ x\equiv y\equiv 0\; (mod\; 13)$. Method 1. One may use Brahmagupta-Fibonacci Identity: \[ \left( A^{2} + B^{2}\right)\left( X^{2} + Y^{2}\right) = \left( AX + BY\right)^{2} + \left( AY - BX\right)^{2}. \] Indeed, we obtain \[ \left({15}^{2} + {16}^{2}\right)\left( a^{2} + b^{2}\right) = \left( 15a + 16b\right)^{2} + \left( 16a - 15b\right)^{2} = x^{4} + y^{4}. \] Since $ {15}^{2} + {16}^{2} = 13\cdot 17$, we find that $ x^{4} + y^{4}$ is divisibly by $ 13$. We now apply the following result with $ p = 13$ to conclude that $ x\equiv y\equiv 0\; (mod\; 13)$. Proposition. Let $ p$ be a prime with $ p\equiv 5\; (mod\; 8)$. Suppose that $ a^{4} + b^{4}$ is divisible by $ p$ for some integers $ a$ and $ b$. Then, both $ a$ and $ b$ are divisible by $ p$. Proof. Assume to the contrary that at least one of them are not divisible by $ p$. Since $ p$ divides $ a^{4} + b^{4}$, we see that none of them are divisible by $ p$. Fermat's Little Theorem yields that $ a^{p - 1}\equiv b^{p - 1}\equiv 1\; (mod\; p)$. Since $ p\equiv 5\; (mod\;8)$ or since $ \frac {p - 1}{4}$ is odd, we have $ ( - 1)^{\frac {p - 1}{4}} = - 1$. Since $ p$ divides $ a^{4} + b^{4}$, we obtain \[ a^{4}\equiv - b^{4}\; (mod\; p). \] Raise both sides of the congruence to the power $ \frac {p - 1}{4}$ to obtain \[ a^{p - 1}\equiv ( - 1)^{\frac {p - 1}{4}}b^{p - 1}\equiv - b^{p - 1}\; (mod\; p). \] or \[ 1\equiv a^{p - 1}\equiv - b^{p - 1}\equiv - 1\; (mod\; p). \] This is a contradiction because $ p$ is an odd prime. Therefore, both $ a$ and $ b$ are divisible by $ p$. Method 2. We now work on the field $ \mathbb{Z}/{13\mathbb{Z}}$. (Since $ 13$ is prime, we know that the ring $ \mathbb{Z}/{13\mathbb{Z}}$ is a field. We identify $ \overline{\alpha}\in\mathbb{Z}/{13\mathbb{Z}}$ with $ \alpha$.) The above congruences become \[ 2x^{2} + 3y^{2} = 0\;\;\text{and}\;\; 3x^{2} - 2y^{2} = 0. \] Of course, Gauss Elimination works. However, we offer an alternative way. Observe that \[ \left[\begin{array}{cc}x^{2} & y^{2} \\ - y^{2} & x^{2}\end{array}\right]\left[\begin{array}{c}2 \\ 3\end{array}\right] = \left[\begin{array}{c}0 \\ 0\end{array}\right]. \] This implies that $ \left[\begin{array}{cc}x^{2} & y^{2} \\ - y^{2} & x^{2}\end{array}\right]$ has zero determinant so that $ x^{4} + y^{4} = 0$. We compute \[ x^{12} = \left( x^{4}\right)^{3} = \left( - y^{4}\right)^{3} = - y^{12} \] If $ y\neq 0$, then we also have $ x = 0$. Hence, we obtain $ 1 = x^{12} = - y^{12} = - 1$, which is a contradiction. Therefore, we get $ y = 0$ and so $ x = 0$.

Hi Ideahitme, I'am sorry to say this but it seems to me that there is something wrong in your solution Namely, 15^2 + 16^2 = 481 but 481 = 13 x 37 ( and not as you write 13 x 17) Thus we work modulo 13 and 37. So, in Step 2 we consider 15x^2 + 16 y ^2 = 0 mod 37 and 16x^2 - 15y^2 = 0 mod 37

mathmanman wrote: $ \left(\frac xy \right)^4 = - 1 \mod 13$ and $ \left(\frac xy \right)^4 = - 1 \mod 37.$ It took me really long to realize these were fractions and not L\'egendre symbols. Just posting this for the next person.

\[ 15a + 16b = x^2,16a - 15b = y^2\Rightarrow (15^2 + 16^2)a = 15x^2 + 16y^2 \] Now $ 15^2 + 16^2 = 481 = 13\cdot 37$ so we need the congruences \[ 15x^2 + 16y^2\equiv 0\pmod{p} \] for each of $ p = 13,37$. In each case, if $ p$ divides $ x$, then $ p$ divides $ y$ as well. Otherwise, we get \[ (4yx^{ - 1})^2\equiv - 15\pmod{p} \] We can check (using QR for example, or the long way... ) that -15 is not a square modulo either prime, so we necessarily find that 481 divides both $ x,y$. But in this case, if $ x = 481m,y = 481n$, then there is always an integer solution for $ a,b$, namely \[ a = 481(15m^2 + 16n^2),b = 481(16m^2 - 15n^2) \] In particular, there's a positive integer solution for $ m = n = 1$, that is, $ x = y = 481$, for $ a = 31\cdot 481,b = 481$. Since $ x,y$ positive and divisible by 481, the minimum is $ \boxed{481}$.

Let $15a+16b=m^2$ and $16a-15b=n^2.$ Sum up their squares to get $$m^4+n^4=481(a^2+b^2)=13\cdot 37(a^2+b^2).$$So $13, 37 \mid m^4+n^4.$ Claim: $13, 37\mid m+n.$ Proof. Assume the contrary that $13\nmid m+n.$ The case with 37 is similar. So we have $$m^4\equiv -n^4\pmod{13} \implies m^{12}\equiv- n^{12} \pmod{13}.$$By Fermat's we get the contradiction that $2\equiv 0 \pmod{13}.$ $\blacksquare$ Let $m=481x, n=481y.$ So $$a=481(15x^2+16y^2) \text{ and } b=481 (16x^2-15y^2).$$For the least, set $x=y=1,$ to get $$a=14911 \text{ and }b=481 \implies b^2=231361.$$