Darn this problem, or at least my solution, is unfortunate. From $a^2=m_2^2+n_2^2$ there exist integers $k\ge1$ and $u>v\ge1$ such that\[ a=k(u^2+v^2)=m_1^2+n_1^2\]and \[|m_2-n_2|=k|u^2-v^2-2uv|.\]Thus we have \begin{align*} 2k(u^2+v^2)-k^2(u^2-v^2-2uv)^2&=2k(u^2+v^2)-(m_2-n_2)^2\\ &=2(m_1^2+n_1^2)-(m_1-n_1)^2\\ &=(m_1+n_1)^2\\ &=m_1^2+n_1^2+2m_1n_1\\ &=k(u^2+v^2)+2\sqrt{m_1^2(a-m_1^2)}\\ &\ge k(u^2+v^2)+2\sqrt{a-1}. \end{align*}It's simple to check that $5<a<19$ has no solutions, so $a\ge19$. Then \begin{align*} 2k(u^2+v^2)-k^2(u^2-v^2-2uv)^2&\ge k(u^2+v^2)+\sqrt{4a-4}\\ &\ge k(u^2+v^2)+\sqrt{72}\\ &>k(u^2+v^2)+8, \end{align*}and since both sides are integral, \[k(u^2+v^2)\ge k^2(u^2-v^2-2uv)^2+9\ge6k|u^2-v^2-2uv|.\]This gives us two inequalities: \[u^2+v^2\ge6u^2-6v^2-12uv\implies u\le v\left(\frac{6+\sqrt{71}}{5}\right)<3v,\]and \[u^2+v^2\ge-6u^2+6v^2+12uv\implies u\ge v\left(\frac{6+\sqrt{71}}{7}\right)>2v,\]where we have used $8<\sqrt{71}<9$ and $u>v>0$. So $2v+1\le u\le3v-1$ for all $v$. Let \[f(k,u,v)=2k(u^2+v^2)-k^2(u^2-v^2-2uv)^2=(m_1+n_1)^2,\]whence $f(k,u,v)$ is a perfect square. To speed up the checking, we recall that $u,v$ must be of opposite parity. For $k=1$, we first test $1\le v\le6$, from which we find that $f(1,15,6)$ is a perfect square, and $a\le261$. Furthermore, $v\ge7\implies u\ge15$, so $a\le261$ is our best bound for $k=1$; now we consider $k\ge3$. If $k$ is a prime $p\equiv3\pmod4$, then since an even number of $p$'s divide $u^2+v^2$, an odd number of $p$'s divide $a=k(u^2+v^2)=m_1^2+n_1^2$, a contradiction. So $k\ge5$, and because we need to find $a\le260$ now, we must have \[u^2<u^2+v^2=\frac{a}{k}\le\frac{260}{5}=52\implies u<\sqrt{52}<8\implies u\le7.\]From $2v+1\le u\le3v-1$ and $u\le7$, we get that $v=1$ has no solutions, $v=2\implies u=5$, $v=3\implies u=7$, and $4\le v\le6$ has solutions. For the two pairs $(u,v)=(5,2),(7,3)$ we have \[k=\frac{a}{u^2+v^2}\le\frac{260}{5^2+2^2}<9\implies k\le8.\]It remains to check $k=5$, since we've established that $k=3$ and $k=7$ yield contradictions. Since $5(7^2+3^2)=290>260$, we just have to check $f(5,5,2)=265$, which is not a perfect square. Thus $a=261$ is the desired minimal number.

Say $n_{1}>m_{1}$ $ a^{2}=(m_{1}^{2}+n_{1}^{2})^{2}=(m_{1}^{2}- n_{1}^{2})^{2} +(2m_{1}n_{1})^{2} $ $n_{2}=n_{1}^{2}- m_{1}^{2} $ $ m_{2}=2m_{1}n_{1} $ $n_{2}-m_{2}=(n_{1}-m_{1})=(n_{1}-m_{1})(n_{1}+m_{1})-2m_{1}n_{1} $ Let $\gcd(n_{1},m_{1})=d \implies n_{1}=xd$ and $m_{1}=yd$ $(n_{1}-m_{1})\mid 2m_{1}n_{1} \implies (x-y)\mid 2xyd$ And since $\gcd(xy,x-y)=1$ and exactly one of $x$ or $y$ is even $\implies d=x-y$ $d=\frac{2xy+1}{x+y}$ Now, $d=3$ for $x=5$ and $y=2$ $n_{1}=15$ and $m_{1}=6$ And $a=261$

darkside102 wrote: $n_{2}=n_{1}^{2}- m_{1}^{2} $ $ m_{2}=2m_{1}n_{1} $ How do you know that this is optimal? There may be more solutions to $(m_1^2+n_1^2)^2=x^2+y^2$ than just $(n_1^2-m_1^2,2m_1n_1)$. Although I would be very happy if there is a nice way to show this...

My idea of equating $n_{2}=n_{1}^{2}- m_{1}^{2} $ & $ m_{2}=2m_{1}n_{1}$ was to find a bound for a, and then prove that there are no smaller solutions. However it does not seem to be an easy task. So I will try a slightly different approach. Let me know if I'm wrong. From the theory of rational points on a circle, another solution will exist if we can write $a$ in another way as sum of two squares say $a=p_{1}^{2}+q_{1}^{2}$ where $\frac{n_{1}}{p_{1}}\neq\frac{m_{1}}{q_{1}}$ Also for any $k=n_{1}-m_{1}=n_{2}-m_{2}$ there will be unique values of $n_{1},m_{1},n_{2}$ and $m_{2}$ since the straight line $x-y=k$ intersects each of the circles $x^{2}+y^{2}=a$ and $x^{2}+y^{2}=a^{2}$ at only one point in the first quadrant. Say $ n_{2}=(p_{1}^{2}-q_{1}^{2})$ and $m_{2}=2p_{1}q_{1} $ So it remains to check $ n_{2}-m_{2}=(n_{1}-m_{1})=|{(p_{1}^{2}-q_{1}^{2})-2p_{1}q_{1}| }$ Now if $\gcd(n_{1},m_{1})=d$ Then $d\mid p_{1}$ and $d\mid q_{1}\implies d\mid (n_{1}-m_{1})\implies d\mid (x-y) \implies x-y\geq d $ Since we are minimizing $a$ we take $x-y=d$. So if $p_{1}=rd$ and $q_{1}=sd$ we get $r^{2}-s^{2}-2rs=\pm1\implies (r-s)^{2}-2s^{2}=\pm1$. Smallest solution for which $a>5$ is $r-s=3$ and $s=2$. So $p_{1}=5d$ and $q_{1}=2d$ and $a=29d^{2}$. But $d=x-y$,so we take $d=3$. Which gives $a=261$.

darkside102 wrote: Now if $\gcd(n_{1},m_{1})=d$ Then $d\mid p_{1}$ and $d\mid q_{1}\implies d\mid (n_{1}-m_{1})\implies d\mid (x-y) \implies x-y\geq d $ If I correctly understand what you wrote, the issue here is that $d^2|a=n_1^2+m_1^2=p_1^2+q_1^2$ does not necessarily mean that $d|p_1$ and $d|q_1$. For example, $5^2|3^2+4^2$, yet $3$ and $4$ are not divisible by $5$.

math154 wrote: darkside102 wrote: Now if $\gcd(n_{1},m_{1})=d$ Then $d\mid p_{1}$ and $d\mid q_{1}\implies d\mid (n_{1}-m_{1})\implies d\mid (x-y) \implies x-y\geq d $ If I correctly understand what you wrote, the issue here is that $d^2|a=n_1^2+m_1^2=p_1^2+q_1^2$ does not necessarily mean that $d|p_1$ and $d|q_1$. For example, $5^2|3^2+4^2$, yet $3$ and $4$ are not divisible by $5$. No I did not mean that...for example the smallest $a$ which can be represented as a sum of two squares in two ways is I think is $50=5^{2}+5^{2}=7^2+1^2$, here $\gcd(7,1)=1$ and $1$ obviously divides every thing. Similarly $65=7^{2}+4^{2}=8^{2}+1^{2}$ again here d=1 We can always start with d=1 and move upwards.

darkside102 wrote: We can always start with d=1 and move upwards. Sorry, but I still don't really understand what you're trying to do... all I can think of is that you want to reduce the problem to the case $d=1$, which doesn't seem to be the case given what you've written... Also, I don't think you can necessarily set $n_2=p_1^2-q_1^2$ and $m_2=2p_1q_1$. You can only guarantee that $n_2=k(p_1^2-q_1^2)$ and $m_2=k(2p_1q_1)$ for some positive $k$, unless your solution prevents this from happening (but again, I don't really understand it).

OK, so if $a=m_1^2+n_1^2=p_1^2+q_1^2$ and $n_1/p_1\ne m_1/q_1$, then why is $n_2=p_1^2-q_1^2$ and $m_2=2p_1q_1$ optimal? In other words, how do you know for sure that the least odd number $a>5$ satisfying the problem's conditions has $a=m_1^2+n_1^2=k(r_1^2+s_1^2)$, $n_2=k(r_1^2-s_1^2)$, and $m_2=k(2r_1s_1)$ such that $m_2-n_2=m_1-n_1$, but that for this value of $a$, there do not exist $p_1,q_1$ such that $a=m_1^2+n_1^2=p_1^2+q_1^2$, $n_1/p_1\ne m_1/q_1$, $n_2=p_1^2-q_1^2$, $m_2=2p_1q_1$, and $n_2-m_2=n_1-m_1$? Basically I just don't understand your solution at all. Could you please write it more precisely so that it's easier to understand?