Sum of two squares and the rearrangement Let $a,b,c,d$ be four nonzero integers such that $ab=cd$. If $a^2+b^2\ne c^2+d^2$, then what is the minimal value of $|(a^2+b^2)-(c^2+d^2)|$? Surely it must be bigger than or equal to $1$, but I guess it is strictly bigger than $1$. How can I prove(or disprove) this conjecture?

You can write $|a^2+b^2-c^2-d^2|=|a^2+2ab+b^2-c^2-2cd-d^2|=|(a+b)^2-(c+d)^2|$ and note that the difference of squares is never $1$ unless they are $0,1$.  To have them $0,1$ we would need $c=-d, a=1-b$ but then we would need $-c^2=b-b^2$ or $b(b-1)=c^2$, which requires $c=0$  
Added:  WOLOG we can ask $a+b \gt c+d$ and get rid of the absolute value signs.  We can also require that all the variables be positive.  Then $a^2+b^2-c^2-d^2=(a+b+c+d)(a+b-c-d)$ and we note the two factors have the same parity so the product cannot be $2$ or $6$.  If the product is $3$, we must have $c+d=1$, which is not possible.  If the product is $4$, both factors are $2$ and $c+d=0$.  If the product is $5$, we have $c+d=2, c=d=1, ab \neq cd$.  If the product is $7$, $c+d=3, c=2, d=1, ab \neq cd$.  If the product is $8$, the factors must be $4,2$, $c+d=1$, again not possible.  So $9$ is the smallest product and you have found an example.
