# How many ordered pairs of integers $(a, b)$ are needed to guarantee that there are two ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ such that $\dots$

Question:
How many ordered pairs of integers $(a, b)$ are needed to guarantee that there are two ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ such that $a_1 \bmod 5 = a_2 \bmod 5$ and $b_1 \bmod 5 = b_2 \bmod 5$?

This is my first time with pigeon hole principle and this is a book exercise without an answer key. Thus I need someone kindly to point out if anything is wrong with my proof and how to make it more straight forward next time.

My Attempt:
For each $a, b$ in $(a, b)$, there are 5 different congruence classes in $\mod 5$. To answer the question of How many ordered pairs of integers of the form $(a, b)$?, we need to consider 25 different congruence classes that both $a, b$ will form. Thus the problem is now reduced to pigeon hole problem,

$$\lceil \dfrac{n}{25}\rceil = 2$$

By solving the inequality of of the ceil function (I don't have to but just to be formal),

$$\lceil \dfrac{n}{25}\rceil-1 < \dfrac{n}{25} \leq \dfrac{n}{25}$$ $$2-1 < \dfrac{n}{25} \leq 2$$ $$25 < n \leq 50$$

Thus have to be $25+1 = 26$ by the Pigeon hole principle.

Yes, the answer is correct. There are $5$ options for the congruence of $a$ and $5$ options for the congruence of $b$ modulo $5$ so there are $25$ options for the congruence of the pair $(a,b)$. Clearly if you have $26$ then at least two will be in the same "congruence pair class". However with $25$ it could be we had one pair of each class, so the answer is 26 as you claim.