How to show that a large given integer isn't a square number by using mod?

Hint. The squares modulo 11 are $0,1,4,9,5,3$. Since $10^k\equiv (-1)^k \pmod{11}$ it follows that the number $4531893869$ modulo $11$ can be obtained by computing: $$-4+5-3+1-8+9-3+8-6+9\pmod{11}.$$ If it does not belong to $\{0,1,4,9,5,3\}$ then $4531893869$ can not be a square modulo 11 and therefore it is not a perfect square.

The key point is that not every value mod $11$ can be a square. In fact for any square $n^2$, $n$ must be congruent to one of $0,1,...,10$ mod $11$, and so $n^2$ must be congruent to one of $0^2,1^2,...,10^2$ mod $11$. So as long as none of $0^2,1^2,...,10^2$ is congruent to $8$ mod $11$, there are no squares congruent to $8$ mod $11$, and your number can't be square.

In fact you can save yourself some work here: $10\equiv -1$ mod $11$, so $10^2\equiv(-1)^2=1^2$, so once you've checked $1^2$ you don't need to check $10^2$. Similarly with other pairs of numbers which have sum $11$, so in fact it is sufficient to check none of $0^2,1^2,...,5^2$ is congruent to $8$ mod $11$.

Hint $\ {\rm mod}\ 11\!:\ n^{\large 2}\equiv 2^{\large 3}\equiv x\ $ when raised to $\color{#c00}5$'th power contradicts little Fermat, viz.

$\quad\overset{(\ \ \ )^{\Large\color{#c00} 5}}\Longrightarrow\,\ \underbrace{ 1\equiv (n^{\large 2})^{\large\color{#c00} 5}}_{\rm Fermat}\!\equiv (2^{\large 3})^{\large\color{#c00} 5}\equiv (2^{\large 5})^{\large 3}\equiv (-1)^{\large 3}\equiv -1,\ $ but $\,\ 1\not\equiv -1\pmod{\!11}$

Remark $ $ This method of square testing works generally - see Euler's criterion. It is much more efficient than brute-force enumerative tests for larger numbers.

Generally we can disprove equalities of integer arithmetical expressions by checking that they are not congruent modulo $m$. This works because modular reduction is compatible with addition and multiplication (see the Congruence Sum and Product Rules), so it preserves equalities (as congruences) between integer expressions composed of sums and products, i.e. polynomial expressions of integers, e.g. $$ P(i,j,k) = Q(i,j,k) \ \Rightarrow\ P(i,j,k)\equiv Q(i,j,k)\pmod m$$

for any polynomials $P,Q$ with integer coefficients. So if the congruence fails for some modulus $m$ then the LHS equality must also fail.