Lets denote $15a+16b=x^2$, $16a-15b=y^2$ so we have $x^4+y^4=(15^2+16^2)(a^2+b^2)$, since $x^4=(15b+16b)^2$,$y^4=(16a-15b)^2$ summing up this two we get desired. So $x^4+y^4=481(a^2+b^2)$ also $481=13\times 37$. We note that $m^4\equiv -1 (mod13)$(*) for m, natural number, but $m^12\equiv1(mod13)$ [by F.L.T] therefore (*) leads to contradiction. Analogously $m^4\equiv-1(mod37)$ gives contradiciton since $m^36\equiv1 (mod37)$. So $x^4+y^4 \equiv0 (mod13)$ states that $x\equiv y\equiv 0 (mod13)$, neither x and y doesnt leave 0 on division by 13. And the last one is impossible, since $m^4+n^4 \equiv 0 (mod13)$ and $m^4$ doesnt leave $-1(mod13)$, if $m^4+n^4\equiv0 (mod13)$ and $n$ is not congruent to $0$ in $mod13$; noting that 13- is prime, we have a natural k, such that $kn\equiv 1(mod13)$, but from here $(mk)^4\equiv -(nk)^4 \equiv -1$. Thus $x\equiv y\equiv0 mod13$ and analogously $x\equiv y\equiv 0mod37$. Resulting we have $x,y\geq481$ and minimal value of $x^2,y^2$ is $481^2$ Answer: $481^2$ davron

Here's a different solution. Put $ 15a + 16b = x^{2}$ and $ 16a - 15b = y^{2}$ as davron did. Then $ 240a + 256b = 16y^{2}$ and $ 240a - 225b = 15y^{2}$; subtracting gives $ 481b = 16x^{2} - 15y^{2}$. Also, $ 225a + 240b = 15x^{2}$ and $ 256a - 240b = 16y^{2}$; adding gives $ 481a = 15x^{2} + 16y^{2}$. Note that $ 481 = 13\cdot37$. So we have $ 37|(16x^{2} - 15y^{2}) - (15x^{2} + 16y^{2}) = x^{2} - 31y^{2}$, so $ x^{2}\equiv31y^{2}\pmod{37}$. Assume that $ y$ is not divisible by 37, so it is relatively prime to 37. Then, $ x$ isn't divisible by 37 either. We have that $ 31y^{2}$ is a quadratic residue modulo 37, but $ \Big(\dfrac{31y^{2}}{37}\Big) = \Big(\dfrac{31}{37}\Big)\cdot\Big(\dfrac{y^{2}}{37}\Big) = \Big(\dfrac{6}{31}\Big)\cdot( - 1)^{15\cdot18}\cdot1 = \Big(\dfrac{2}{31}\Big)\cdot\Big(\dfrac{3}{31}\Big) = ( - 1)^{15\cdot16}\cdot\Big(\dfrac{1}{3}\Big)\cdot( - 1)^{1\cdot15} = - 1$, contradiction. So $ 37|x,y$. We also have $ 13|x^{2} - 31y^{2}$, so $ x^{2}\equiv5y^{2}\pmod{13}$. Assume that $ y$ is not divisible by 13, so it is relatively prime to 13. Then, $ x$ isn't divisible by 13 either. We have that $ 5y^{2}$ is a quadratic residue modulo 13, but $ \Big(\dfrac{5y^{2}}{13}\Big) = \Big(\dfrac{5}{13}\Big)\cdot\Big(\dfrac{y^{2}}{13}\Big) = \Big(\dfrac{3}{5}\Big)\cdot( - 1)^{2\cdot6}\cdot1 = \Big(\dfrac{2}{3}\Big)\cdot( - 1)^{1\cdot2}\cdot1 = - 1$, contradiction. So $ 13|x,y$. Now, $ 481|x,y$, and since 481 is square-free, $ 481^{2}|x^{2},y^{2}$, and $ x^{2}$ and $ y^{2}$ are at least $ 481^{2}$. To get equality, take $ a=481\cdot31, b=481$.

Yet another solution. The minimum possible value may be attained by setting $a = 13 \cdot 31\cdot 37$ and $b = 13 \cdot 37$, which causes $15a + 16b = 16a - 15b =481^2$. We prove this below. Let $15a + 16b = m^2$ and $16a - 15b = n^2$ for integral $m,n$. Solving for $a$ and $b$ in terms of $n$, we get: \[ a=\frac{16m^2 - 15n^2}{13\cdot 37},b=\frac{15m^2 + 16n^2}{13\cdot 37} \] Since $a$ is integral, we must have that $16m^2 \equiv 15n^2 \bmod 13\Rightarrow m^2 \equiv 5n^2 \bmod 13$. The quadratic residues, modulo $13$ are $0$, $\pm 1$, $\pm3$, $\pm 4$. Multiplying these residues by $5$ yields $0$, $\pm 2$, $\pm 5$, $\pm 7$, modulo $13$, suggesting that $13 \mid m, n$. Further, we must have that $16m^2 \equiv 15n^2 \bmod 37 \Rightarrow m^2 \equiv 31n^2 \bmod 37$. The quadratic residues, modulo$37$ are $0$, $\pm 1$, $\pm 3$, $\pm 4$, $\pm 7$, $9$, $\pm 10$, $\pm 11$, $\pm 12$, $\pm 16$. Multiplying these residues by $31$ yields $0$, $\pm 2$, $\pm 5$, $\pm 6$, $\pm 8$, $\pm 13$, $\pm 14$, $\pm 15$, $\pm 18$, $20$, modulo $37$, suggesting that $37 \mid m, n$. Letting $m = 13 \cdot 37 \cdot m_1, n = 13 \cdot 37 \cdot n_1$ and substituting, we get that $a = 13\cdot 37 \cdot (15m_1^2 + 16n_1^2)$ and $b = 13\cdot 37 \cdot (16m_1^2 - 15n_1^2)$. Substituting once more, our we get that $15a + 16b = 13^237^2m_1^2$ and $16a - 15b = 13^237^2n_1^2$. Setting $m_1 = n_1 = 1$ yields the desired result.

Check here for my solution and some others.

The answer is $\boxed{481^2}$, the construction being $b = 481, a = 31 \cdot 481$. Let $m^2 = 15a + 16b, n^2 = 16a - 15b$ for positive integers $m,n$. Note that $16m^2 - 15n^2 = 481b = 13\cdot 37 \cdot b$. Claim: $13\mid m$ Proof: Suppose not. Then $13\nmid n$ also. Thus, $16m^2 \equiv 15 n^2 \pmod{13}$, so $\frac{16}{15}$ must be a quadratic residue modulo $13$. But this means that $\frac 32$ must be a QR modulo $13$. But, $\frac 32 \equiv 8 \pmod{13}$, so we get that $2$ must be a QR mod $13$, but this is false since $13$ isn't $-1$ or $1$ modulo $8$. $\square$ Claim: $37 \mid m$. Proof: Suppose not. Then $37\mid n$ also. Thus, $16m^2 \equiv 15n^2 \pmod{37}$, so $\frac{16}{15}$ is a QR modulo $37$. But, $\frac{16}{15} \equiv \frac{37 \cdot 2 + 16}{15} \equiv 6 \pmod{37}$. Now, $3$ is a QR mod $37$ since $37 \equiv 1\pmod{12}$, but $2$ isn't since $37 \equiv 5 \pmod 8$. Hence $6$ isn't a QR mod $37$, a contradiction. $\square$ Hence $481 \mid m$, and therefore $481\mid n$ also. This proves that $\min(m^2, n^2) \ge 481^2$, as desired.

Let $15a+16b=x^2$, $16a-15b=y^2$. This implies $a=\frac{15x^2+16y^2}{481}$. Note that $481=13\cdot 37$. Now, we will show $x,y$ are divisible by $481$. If $x,y$ aren't divisible by $13$ then, $(x/y)^2 \equiv -16/15 \equiv 5 \pmod{13}$, a contradiction since $5$ isn't a quadratic residue $\pmod{13}$. If $x,y$ aren't divisible by $37$ then $(x/y)^2 \equiv -16/15 \equiv -6 \pmod{37}$. But using Euler's criterion we find $$\left (\frac{-6}{37} \right)=\left (\frac{-1}{37} \right)\left (\frac{2}{37} \right)\left (\frac{3}{37} \right)=1\cdot (-1)\cdot 1=-1,$$a contradiction. Hence $x,y$ are divisible by $481$. So the least possible value of $y^2$ is $481^2=231361$, which works.