China Girls' Mathematical Olympiad

The answer is $261$.

Note that $$261=15^2+6^2,261^2=189^2+180^2,15-6=189-180.$$ We know that there is no number in between $5$ and $261$ that satisfies the condition of the problem. Assume on the contrary that $a$ is such a number. We may set $d=m_1-n_1>0$. Because $a$ is odd, $m_1$ and $n_1$ have different parity, and so $d$ is odd. Because $m_1<261,d\le 15$; that is, the possible values of $d$ are

$1,3,5,7,9,11,13,15$. We will eliminate every one of them.

We can write $m_2=n_2+d$ and $a^2=(n_2+d{)}^2+n_2^2$ or $$2a^2-d^2=(2n_2+d{)}^2.\stackrel{◯}{1}$$ If $d=1$, then $\stackrel{◯}{1}$ becomes a Pell's equation $x^2-2y^2=-1$ with $(x,y)=(2n_2+1,a)$. This Pell's equation has minimal solution $(x,y)=(1,1)$ and $x+y\sqrt{2}=(1+\sqrt{2}{)}^{2k-1}$ for positive integers $k$. The $y$ values of the solutions of this Pell's equation are $5,29,169,985,\dots$. Thus, the only possible values for $a$ are $29$ and $169$. It is easy to check that neither $29$ nor $169$ can be written in the form of $(n_1+1{)}^2+n_1^2$. Hence $d\ne 1$.

If $d$ is a multiple of $3$, then $m_2\equiv n_2(\mathrm{mod}3)$ and $a^2\equiv 2m_2^2(\mathrm{mod}3)$. Because $2$ is not a quadratic residue modulo $3$, we conclude that $0\equiv m_2\equiv n_2\equiv a(\mathrm{mod}3)$. Hence, $m_1^2+n_1^2\equiv 0(\mathrm{mod}3)$, from which it follows that $m_1\equiv n_1\equiv 0(\mathrm{mod}3)$. Thus, $a=m_1^2+n_1^2$ is a multiple of $9$ and $m_2^2+n_2^2=a^2$ is a multiple of $81$. We can write $m_2=3m^{\prime }$, $n_2=3n^{\prime }$, and $a=9a^{\prime }$ for integers $m^{\prime },n^{\prime },a^{\prime }$. We have $m^{\prime 2}+n^{\prime 2}=9a^{\prime 2}$. Again, because $-1$ is not a quadratic residue modulo $3$, we must have $m^{\prime }\equiv n^{\prime }\equiv 0(\mathrm{mod}3)$. It follows that $d=3(m^{\prime }-n^{\prime })$ is divisible by $9$. Thus, $d=9$.

Because $a<261=15^2+6^2$, $n_1<6$. Therefore, $n_1=3$ and $a=12^2+3^2=153$. But then $\stackrel{◯}{1}$ becomes $9^2\cdot 577=(2n_2+9{)}^2$, which is impossible. Hence, $d\ne 3$, or $9$, or $15$.

If $d=11$ or $d=13$, because $2$ is not a quadratic residue modulo $d$, from $a^2\equiv 2m_2^2(\mathrm{mod}d)$, we conclude that $0\equiv m_2\equiv n_2\equiv a(\mathrm{mod}d)$. It follows that $2m_1^2\equiv a\equiv 0(\mathrm{mod}d)$ and $0\equiv m_1\equiv n_1\equiv a(\mathrm{mod}d)$. In particular, $n_1\ge d$, $m_1\ge 2d$ and $$a=m_1^2+n_1^2\ge 5d^2>261.$$ Hence, $d\ne 11$ or $13$.

If $d=5$, we also note that $2$ is not a quadratic residue modulo $5$. By the same reasoning before, we have $$m_1\equiv n_1\equiv m_2\equiv n_2\equiv 0(\mathrm{mod}5).$$ If $n_1\ge 10$, then $m_1\ge 15$ and $a=m_1^2+n_1^2\ge 10^2+15^2>261$. Thus, $n_1=5$, $m_1=10$, and $a=125$. But then $\stackrel{◯}{1}$ becomes $$5^2\cdot 1249=(2n_2+5{)}^2,$$ which is impossible. Hence, $d\ne 5$.

If $d=7$, then because $a<261=15^2+6^2$, we have $n_1\le 8$. The possible values of $a$ are then $65,85,109,137,169,205,245$. But then $\stackrel{◯}{1}$ becomes $(2n_2+7{)}^2=2a^2-49$. It is easy to check that there is no solution in this case. Hence, $d\ne 7$.

Combining the above, we conclude that $261$ is the answer of this question. $\square$