2

Question: Does the sequence 144; 1,444; 14,444; 144,444; ... contain any squares besides 144 and 1,444?

One can easily see that $12^2=144$ and $38^2=1,444$, so I am wondering whether the sequence contains any further squares.

The number 14,444 is not a square, because it is strictly between $120^2=14,400$ and $121^2=14,641$ (found in row 4 in Pascal's triangle).

Similarly, the number 144,444 is not a square either, because it is strictly between $380^2=144,400$ and $381^2=145,161$.

More generally, as one might expect, the square root of one followed by an even number of fours begins with the digits one and two, while the square root of one followed by an odd number of fours begins with the digits three and eight.

Using modular arithmetic, one can easily show an integer whose square ends with a four in base ten must either end with a two or an eight in base ten.

Geoffrey Trang
  • 10,823

3 Answers3

2

You can prove no square ends in at least four $4$s. If it did, we could halve the choice of the square root to get a square ending in $11$, which would leave a remainder of $3$ when divided by $4$.

J.G.
  • 118,053
1

This is a clarification of the answer posted here by J.G.

$11$ is not a perfect square modulo $100$, since $3$ is not a perfect square modulo $4$ (the only perfect squares modulo $4$ are $0$ and $1$).

Now $44$ is a perfect square modulo $100$ (it is in the same residue class as $12^2$) but $4444$ is not a perfect square modulo $10000$, since after dividing by $4$ we obtain either $1111, 3611, 6111,$ or $7611$ mod $10000$ and all of these belong to the residue class of $11$ mod $100$, which we already explained is not a perfect square.

pre-kidney
  • 30,884
0

$4444$ is not a square in $\mathbb{Z}/(5^4\mathbb{Z})^*$. Indeed $a^2\equiv 4444\pmod{5^4}$ implies $a^2\equiv 444\pmod{5^3}$, hence $a=125k\pm 38$, but no $k\in\{0,1,2,3,4\}$ is such that $(125k\pm 38)^2\equiv 4444\pmod{5^4}$. It follows that the only squares in such sequence are $144$ and $1444$.

Jack D'Aurizio
  • 361,689