An algebra problem by Dhruv Sharma

$a+1=b+2 \implies a-1=b$ $a+1=c+3 \implies a-2=c$ $a+1=d+4 \implies a-3=d$ $a+1=e+5 \implies a-4=e$

Adding the above results gives $a+b+c+d+e=5a-10$

Since $a+1=a+b+c+d+e+3$ , we have $a+1=5a-7 \implies a=2 \implies b=1, c=0, d=-1, e=-2.$

Therefore, $a^2+b^2+c^2+d^2+e^2-10=(2)^2+(1)^2+(0)^2+(-1)^2+(-2)^2-10=\boxed{0}$