Okay, I will post my soln, since I think this is a very nice problem. From $a-d > c-b $ we obtain that $ ( a-d)^2 > (b-c)^2 \Rightarrow (a-d)^2+4ad > (c-b)^2 + 4bc $ and thus $(a+d)^2 > (b+c)^2 \Rightarrow a+d \geq b+c+1. $ Suppose first that $\sqrt d - \sqrt a = 1$. Then $ a+d - 2\sqrt {ad} = 1 \Leftrightarrow $ $ a+d = 1 + 2 \sqrt {bc } \leq 1 + b+c \Rightarrow b=c$ and $a+d=2b+1$. But then $ad=b^2$, so $\gcd (a,d)=1$ because otherwise, if $p\ | \ \gcd(a,d)$ and $p$ is a prime we would have $ p\ | \ b$ and thus $p \not | 2b+1$ contradiction. So, each of $a$ and $d$ are perfect squares. Now, suppose that $\sqrt d - \sqrt a <1$. Then $a+d < 1+ 2\sqrt {ad} \leq 1+b+c \leq a+d $ which is a contradiction.

Where do we get thata-d>c-b,and also a-d is negative,but c-b is positive! and we can not square both sides because a-d is negative!

Sorry I ment $|a-d| > |b-c|$ which is obvious given that $a<b\leq c<d$.

spider_boy wrote: Where do we get thata-d>c-b,and also a-d is negative,but c-b is positive! and we can not square both sides because a-d is negative! The solution still works after squaring regardless of sign because of absolute value, but yes that is an error. Nice problem!