Hint $\rm\, 1575 = 15\!\cdot\! 100\! +\! 75 = 25\,(15\!\cdot\! 4\! +\! 3) = 5^2\cdot 3^2\cdot\color{#C00} 7\:$ lacks only one $\,\color{#C00}{??}\,$ to become square.
Remark $\ $ Suppose, instead, that $\rm\:n\:$ is not $\,1575\,$ but is a bigger integer that is difficult to factor completely, but we are given that it is not square. First, we can rule out $\rm\:k = 9\,$ and $\,25,\,$ since multiplying $\rm\:n\:$ by a square does not change the squareness of $\rm\:n.\:$ Since $\,63 = 7\cdot 3^2,\:$ we infer that if $\rm\:63\,n = 3^2\!\cdot\! 7\,n\:$ is a square then so too is the smaller $\rm\:7n.\:$ This leaves only the possibilities $\rm\: k = 7\:$ or $\rm\:15 = 3\!\cdot\! 5.\:$ To determine which $\rm\:kn\:$ is square, we need only determine the parity of the power of any one of the primes $\,3,5,7\,$ in the factorization of $\rm\:n.\:$ For example, in your case, once we have determined that $\rm\:n = 5^2\,j,\ 5\nmid j,\:$ then we know $\rm\:k\:$ must have have an even power of $5$, which excludes $\rm\:k=3\!\cdot\!5,\:$ leaving $\rm\:k = 7\:$ as the only possible solution.