If  $(a-1)^2+(b-2)^2+(c-3)^2+(d-4)^2=0$ , then what is the value of the expression  $abcd+1$.

Factorize this expression using the difference of squares: 

$W+X+Y+Z=0$

Now, if all of W, X, Y and Z are complete squares, which they are according to the question statement, and their sum is 0, we can be sure that all of them are 0 as well, or mathematically put,

$W=0,\ X=0,\ Y=0,\ Z=0,$ if $W+X+Y+Z=0$ and $W,\ X,\ Y\ and\ Z$ are perfect squares.

According to the given question,

$W=(a−1)2;\ X=(b−2)2;\ Y=(c−3)2;\ Z=(d−4)2$

Thus,

$W=0\Rightarrow (a−1)^2=0\Rightarrow a−1=0 \Rightarrow a=1$

$X=0\Rightarrow  (b−2)^2=0\Rightarrow b−2=0 \Rightarrow b=2$

$Y=0\Rightarrow (c−3)^2=0\Rightarrow c−3=0 \Rightarrow c=3$

$Z=0\Rightarrow (d−4)^2=0\Rightarrow d−4=0 \Rightarrow d=4$

Thus, to answer the question, the value of the expression

$abcd+1=1\times2\times3\times4+1=24+1=25$