How find all this positive of $n$? Let $n$ be a number of the form $a^2+b^2$ with $a,b\in N^{+},(a,b)=1$, such that
every prime  $P \lt \sqrt{n}$ satisfies $P|ab$. Find all such positive integers $n$?
I find $n=5$, and $n=13$,are there any others?
when $n=5$,then $a=2,b=1$, and $P=2<\sqrt{5}$
when $n=13$,then $a=3,b=2$,and  $P=2$ or $3$

First note that $\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \mid ab$. Thus $n=a^2+b^2 \geq 2ab \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p}$.
If $n \geq 26$, then $\sqrt{n}>5$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 5, p \text{prime}}{p}=60$. This implies that $\sqrt{n}>7$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 7, p \text{prime}}{p}=420$. This implies that $\sqrt{n}>19$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 19, p \text{prime}}{p}=199399380$. This implies that $\sqrt{n} \geq \sqrt{199399380}>\sqrt{100000000}=10000$.
Let $k=\lfloor \frac{\sqrt{n}}{16} \rfloor$, so $624=\frac{10000}{16}-1 \leq \frac{\sqrt{n}}{16}-1<k \leq \frac{\sqrt{n}}{16}$. We have $16k \leq \sqrt{n}$. Also, $\sqrt{n}<16(k+1)$ so $n<256(k+1)^2$.
By Bertrand's postulate, there exists primes $p, q, r, s$ with $k<p<2k<q<4k<r<8k<s<16k \leq \sqrt{n}$. Thus $256(k+1)^2 \geq n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2pqrs>2(k)(2k)(4k)(8k)=128k^4>64k^4$. Thus $4(k+1)^2>k^4$, so $2(k+1)>k^2$ so $k \leq 2$, a contradiction.
Therefore $n \leq 25$. Note that $25 \geq n=a^2+b^2>\max(a^2, b^2)$, so $a, b \leq 4$. If $n \geq 10$, then $\sqrt{n}>3$ so $2(3) \mid ab$. WLOG assume $3 \mid b$, so that $b=3$, so $2 \mid a$, so $a=2, 4$. This gives $n=2^2+3^2=13$ and $n=4^2+3^2=25$, which are both solutions. 
Otherwise $n \leq 9$, so $9 \geq n=a^2+b^2>\max(a^2, b^2)$, so $a, b \leq 2$. $a, b$ cannot be both $2$, so we have $n=1^2+2^2=5$ and $n=1^2+1^2=2$, which are both solutions.
In conclusion, all solutions are given by $n=2=1^2+1^2, n=5=1^2+2^2, n=13=2^2+3^2, n=25=4^2+3^2$.
