35. Find the largest integer $N$ such that both $N+496$ and $N+224$ are perfect squares.
35. Ans: 4265 .
Let $N+496=a^{2}$ and $N+224=b^{2}$ for some positive integers $a$ and $b$. Then $a^{2}-b^{2}=496=$ $224=272=2^{4} \cdot 17$. Thus $17 \mid(a+b)(a-b)$. If $17 \mid a-b$ then $a-b \geq 17$ and $a+b \leq 16$, impossible. Thus $17 \mid a+b$.
We have five possibilities for $(a+b, a-b)$ : $(17,16),(34,8),(68,4),(136,2),(272,1)$. Solve and discard non-integer solutions, we have $(a, b)=(21,13),(36,32)$ and $(69,67)$. Thus the largest $N$ is $69^{2}-496=4265$.