Romanian Mathematical Olympiad

a) The next square after $a^2$ is $(a+1{)}^2=a^2+2a+1$, therefore $a^2+1,a^2+2,\dots ,a^2+2a$ fit between two consecutive squares, hence they are not squares.

b) Suppose, by way of contradiction, that such numbers exist. Then $m^2+n+p>m^2$, hence $m^2+n+p\ge m^2+2m+1$. Writing the other two analogous inequalities and adding up yields $$m^2+n^2+p^2+2m+2n+2p\ge m^2+n^2+p^2+2m+2n+2p+3,$$ a contradiction.