18. The sum of the squares of four consecutive positive integers is equal to the sum of the squares of the next three consecutive integers. What is the square of the smallest of these integers?
18. 441 Let the smallest number be $x$. Therefore $x^{2}+(x+1)^{2}+(x+2)^{2}+(x+3)^{2}=(x+4)^{2}+(x+5)^{2}+(x+6)^{2}$ and hence $x^{2}+x^{2}+2 x+1+x^{2}+4 x+4+x^{2}+6 x+9=x^{2}+8 x+16+x^{2}+10 x+25+x^{2}+12 x+36$. This can be rewritten as $4 x^{2}+12 x+14=3 x^{2}+30 x+77$ or $x^{2}-18 x-63=0$. Hence $(x-21)(x+3)=0$, which has solutions $x=21$ and $x=-3$. The question tells us that $x$ is positive and therefore $x=21$.
The square of the smallest of these integers is therefore $21^{2}=441$.