Integers whose Squares end in 444

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Theorem

The sequence of positive integers whose square ends in $444$ begins:

$38, 462, 538, 962, 1038, 1462, 1538, 1962, 2038, 2462, 2538, 2962, 3038, 3462, \ldots$

This sequence is A039685 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

\(\ds 38^2\) \(=\) \(\ds 1444\)
\(\ds 462^2\) \(=\) \(\ds 213 \, 444\)
\(\ds 538^2\) \(=\) \(\ds 289 \, 444\)
\(\ds 962^2\) \(=\) \(\ds 925 \, 444\)
\(\ds 1038^2\) \(=\) \(\ds 1 \, 077 \, 444\)
\(\ds 1462^2\) \(=\) \(\ds 2 \, 137 \, 444\)
\(\ds 1538^2\) \(=\) \(\ds 2 \, 365 \, 444\)
\(\ds 1962^2\) \(=\) \(\ds 3 \, 849 \, 444\)
\(\ds 2038^2\) \(=\) \(\ds 4 \, 153 \, 444\)
\(\ds 2462^2\) \(=\) \(\ds 6 \, 061 \, 444\)
\(\ds 2538^2\) \(=\) \(\ds 6 \, 441 \, 444\)
\(\ds 2962^2\) \(=\) \(\ds 8 \, 773 \, 444\)
\(\ds 3038^2\) \(=\) \(\ds 9 \, 229 \, 444\)
\(\ds 3462^2\) \(=\) \(\ds 11 \, 985 \, 444\)


All such $n$ are of the form $500 m + 38$ or $500 m - 38$:

\(\ds \paren {500 m + 38}^2\) \(=\) \(\ds 250 \, 000 m^2 + 38 \, 000 m + 1444\)
\(\ds \) \(=\) \(\ds \paren {250 m^2 + 38 m + 1} + 444\)


\(\ds \paren {500 m - 38}^2\) \(=\) \(\ds 250 \, 000 m^2 - 38 \, 000 m + 1444\)
\(\ds \) \(=\) \(\ds \paren {250 m^2 - 38 m + 1} + 444\)

and it is seen that all such numbers end in $444$.


Now we show that all such numbers are so expressed.

In Squares Ending in Repeated Digits, we have shown the only numbers with squares ending in $44$ ends in:

$12, 38, 62, 88$

hence any number with square ending in $444$ must also end in those numbers.


Suppose $\sqbrk {axy}^2 \equiv 444 \pmod {1000}$, where $a < 10$ and $\sqbrk {xy}$ is in the above list.

For $\sqbrk {xy} = 12$:

\(\ds 444\) \(=\) \(\ds \paren {100 a + 12}^2\) \(\ds \pmod {200}\)
\(\ds \) \(=\) \(\ds 10000 a^2 + 2400 a + 144\) \(\ds \pmod {200}\)
\(\ds \) \(=\) \(\ds 144\) \(\ds \pmod {200}\)

This is a contradiction.


Similarly for $\sqbrk {xy} = 88$:

\(\ds 444\) \(=\) \(\ds \paren {100 a + 88}^2\) \(\ds \pmod {200}\)
\(\ds \) \(=\) \(\ds 10000 a^2 + 17600 a + 7744\) \(\ds \pmod {200}\)
\(\ds \) \(=\) \(\ds 144\) \(\ds \pmod {200}\)

Again, a contradiction.


For $\sqbrk {xy} = 38$:

\(\ds 444\) \(=\) \(\ds \paren {100 a + 38}^2\) \(\ds \pmod {1000}\)
\(\ds \) \(=\) \(\ds 10000 a^2 + 7600 a + 1444\) \(\ds \pmod {1000}\)
\(\ds \) \(=\) \(\ds 600 a + 444\) \(\ds \pmod {1000}\)

The solutions to $600 a \equiv 0 \pmod {1000}$ are $a = 0$ or $5$.

Hence:

$\paren {500 n + 38}^2 \equiv 444 \pmod {1000}$


Similarly, for $\sqbrk {xy} = 62$:

\(\ds 444\) \(=\) \(\ds \paren {100 a + 62}^2\) \(\ds \pmod {1000}\)
\(\ds \) \(=\) \(\ds 10000 a^2 + 12400 a + 3844\) \(\ds \pmod {1000}\)
\(\ds \) \(=\) \(\ds 400 a + 844\) \(\ds \pmod {1000}\)

The solutions to $400 a + 400 \equiv 0 \pmod {1000}$ are $a = 4$ or $9$.

Hence:

$\paren {500 n + 462}^2 \equiv \paren {500 n - 38}^2 \equiv 444 \pmod {1000}$

$\blacksquare$


Sources

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