Proving Natural Numbers Made of 6’s and 0’s Are Not Squares – Elementary Number Theory

elementary-number-theory

I proceeded using infinite descent. Let $ N =a_na_{n-1}a_{n-2}\ldots\ldots a_2a_1a_0$ be the decimal representation of the number. Then either $N$ ends in an even number of zeroes or $a_0=6$

Now all squares are $\equiv 0 \text{ or } 1 \bmod 4 $. But if $N$ ends in $06\text{ or }66 $, then $N\equiv 2 \bmod 4 $. Thus $ N $ does not end in $6$.

If $N$ ends in an even number of zeroes then, then $ N=10^{2n}\cdot N'$.

Applying the same argument to $ N'$ we commence an infinite descent.

Is the proof correct ? Any shorter proofs ?

Best Answer

To make the last part of the argument of @lsp explict, if a square $x^2$ ends in $6$, then the last two decimal digits of $x$ are

  • either $a4$, and then $$ x^2 \equiv (a \cdot 10 + 4)^2 \equiv a \cdot 80 + 16 \pmod{100}, $$
  • or $a6$, and then $$ x^2 \equiv (a \cdot 10 + 6)^2 \equiv a \cdot 120 + 36 \pmod{100}. $$

In both cases the last-but-one digit is indeed odd.

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