Find the second smallest integer such that its square's last two digits are $ 44 $

Question

Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property.

My approach is two solve the equation $ n^2 \equiv 44 \pmod{100}, $ but I do not know how to proceed to solve that equation.

I try a different path by letting $ n = 10x + y $ for some integers $ x, y, $ where $ 0 \le y \le9. $ Then $ n^2 \equiv 44 \; \pmod{100} $ can be reduced to $ 20xy + y^2 \equiv 44 \pmod{100}. $ At this point I let $ x $ run from $ 0, 1, 2, \dots $ and find the integer $ y \in \mathbb{Z}_{100} $ such that $ y^2 + 20xy - 44 = 0. $

My question is is there an alternative way to tackle this problem without having to try each $ x $ and $ y? $ Maybe try to solve the initial congruence equation $ n^2 \equiv 44 \pmod{100}. $

Answer 1

We can use your idea: Let $n=10x+y$. If the last digit of $n^2$ is $4$ then the last digit of $n$ is either $2$ or $8$, so $y=2$ ot $y=8$.

First let's assume $y=2$. We then have $40x+4\equiv 44\mod100$, or equivalently $40x+4=44+100t$ for some $t$. For $t=0$ we get $x=1$, so $n=12$, and this is not what we want. For $t<0$ we get $x<0$ and these are not solutions. For $t=1$ we get $x=3.5$, not a solution and for $t=2$ we get $x=6$, so $n=62$ is a possible solution.

Now let's assume $y=8$, so, just as above, we need $160x+64=44+100t$ for some $t$, or equivalently $8x+1=5t$, for some $t,x$ intergers. The smallest positive integer solution is $x=3$, $t=5$, and then $n=38$ is a solution for our problem.

$n=38$ is therefore the second smallest positive integer with square ending in $44$.

Answer 2

Hint: If the last two digits of your number are $a$ and $b$, writing down the multiplication (as in primary school) will show you that what we want is

$$ b^{2} \equiv 4 \pmod{10} \quad \text{and} \quad 2ab \equiv 4 \pmod{10} $$

where $a$ and $b$ are digits (between 0 and 9).

Answer 3

$n^2 \equiv 44 \pmod {100}$

By Chinese remainder theorem, $n^2 \equiv 0 \pmod 4$ and $n^2 \equiv 19 \pmod {25}$ is an solution for the formula above.

$$n^2 \equiv 0 \pmod 4 \tag{1}$$ $$n^2 \equiv 19 \pmod {25} \tag{2}$$

By (1), $n = 2k$. Substitude into (2),

$$k^2 \equiv 19 \times 4^{-1} \equiv 19 \times 19 \equiv 11 \pmod {25} \tag{3}$$

By (3), $k \equiv 6 \pmod {25}$ or $k \equiv 19 \pmod {25}$,

when $k = 25w +6$, $n = 50w + 12$.

when $k = 25w +19$, $n = 50w + 38$.

Addendum: The procedure to get the inverse 4 under module 25 is same to get quadratic residue for 11 under module 25.

Here is the manifestation for the latter one.

Since $n^2 \equiv 11 \equiv 1 \pmod 5$, by numerating, $n \equiv 1,4 \pmod 5$.

When $n = 5k+1$, $n \equiv 1, 6, 11, 16, 21 \pmod {25}$

Check one after another and eliminate, gets $n \equiv 6 \pmod{25}$.

When $n = 5k+4$, same procedure plays.

So $k \equiv 6,19 \pmod{25}$ such that $k^2 \equiv 11 \pmod{25}$

Answer 4

If $x^2$ ends with $44$ then $x$ is even. Let $y=2x$. We are trying to solve $$(2y)^2\equiv 44\pmod{100}$$ and this equation is equivalent to $$y^2\equiv 11\pmod{25}$$ Since $6^2\equiv 11\pmod{25}$ this equation can be written as $$(y-6)(y+6)\equiv 0\pmod{25}$$ It is not possible that both $y-6$ and $y+6$ are multiples of $5$, so the solutions are $y\equiv\pm6\pmod{25}$. The first positive values for $y$ are $$6,19,31,44,56,69,81,94,106,\ldots$$ Multiply these numbers by $2$ to get the solutions for $x$.