If $a<b<c<d$ are distinct positive integers such that $a+b+c+d$ is a square, what is the minimum value of $c+d$ ?
The answer is 11 , which comes from $2+3+5+6$ or $1+4+5+6$.

The minimum possible value of $a+b+c+d$ for $a<b<c<d$ is $1+2+3+4=10$. The smallest square at least 10 is $4^{2}$.

Now assume for sake of contradiction that $c+d \leq 10$. Then, $c<\frac{c+d}{2} \leq 5$, so $a<b<$ $c \leq 4$ and $a+b \leq 2+3=5$. Therefore, $16 \leq a+b+c+d=(a+b)+(c+d) \leq 5+10=15$, a contradiction, confirming that 11 is minimal.