A Gardner-esque puzzle

One of my favourite sources of puzzles at the moment is @WWMGT - What Would Martin Gardner Tweet? (Martin Gardner, in case you’re not up on the greats of popular maths writing, was one of the greats of popular maths writing - and is indirectly responsible for Big MathsJam.)

Recently, it was decreed that Martin Gardner would have tweeted:

Show that no square of two or more digits can have only odd digits.

Must be easy, I thought. Let’s do it by contradiction, and try to find a square number - $k^{2}$ with only odd digits.

$k$ has to be odd, because otherwise it would end in an even digit - so it must end in 1, 3, 5, 7 or 9.

If $k$ ends in 1, it can be written as $k = 10 n + 1$ , so $k^{2} = 100 n^{2} + 20 n + 1$ ; its last digit is 1, but the one before it is even ( $\frac{k^{2} - 1}{10} = 10 n^{2} + 2 n$ ).

A similar argument accounts for $k$ ending in 3: if $k = 10 n + 3$ , $k^{2} = 100 n^{2} + 60 n + 9$ , which again has an even penultimate digit.

It was at exactly this point that inspiration struck.

You see, I’d been worried about how I was going to deal with awkward numbers like somethingty-seven and somethingty-nine - but then it struck me: I can write those as $10 n - 3$ and $10 n - 1$ and apply the same reasoning as before! That just leaves me with somethingty-five.

If $k = 10 n + 5$ , then $k^{2} = 100 n^{2} + 100 n + 25$ , in which case the penultimate digit has to be 2.

And we’re done! None of the possibilities hold up, so our assumption that there was such a number must have been mistaken.

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