Do there exist infinitely many solutions to the equation $$\overline{x^2y^2} = z^2$$ where $x,y,z$ are positive integers? Note: the notation $\overline{xy}$ means a concatenation of the numbers $x$ and $y$ to form one number.

One solution to this equation is $(x,y,z) = (4,9,41)$. Then we have $\overline{4^29^2} = 1681 = 41^2$. If $y = \overline{a_1 \ldots a_n}$, then $\overline{x^2y^2} = 10^{2n} x^2+y^2 = z^2$, but I didn't see how to find infinitely many integer solutions to this.

  • 2
    $\begingroup$ It might be useful to explain that the notation (overline) $\overline{x^2y^2} = z^2$ indicates that you are asking whether there are infinitely many solutions to the concatenation of $x^2 $ with $y^2$, in that order, which is equal to square of some integer. $\endgroup$ – amWhy Jul 3 '17 at 22:05
  • $\begingroup$ Thanks for the clarification. I just went to Wikipedia and the Vinculum article doesn't even include concatenation as a possible use. $\endgroup$ – ZTaylor Jul 3 '17 at 22:07
  • $\begingroup$ Also $\,(2,3,7)\,$. $\endgroup$ – dxiv Jul 3 '17 at 22:08
  • $\begingroup$ If we take y=1,then we get 10x^2+1=z^2 which gives z^2-10x^2=1 .Since,this is a positive pell equation and thus have infinitely many solutions. Does this solution work? $\endgroup$ – Yes it's me May 10 '20 at 8:24

$$\overline{2^2(3\cdot10^k)^2}=(7\cdot10^k)^2$$ and in general, if $(a,b,c)$ is a solution, then so is $(a,b\cdot10^k,c\cdot10^k)$

  • $\begingroup$ @KitterCatter Yes; you end up being able to factor out the squared power of $10$ $\endgroup$ – TomGrubb Jul 3 '17 at 22:14

A preliminary report on using an idea that should give us many solutions.

Possibly more interesting (=avoiding trailing zeros in $z$) examples can be found as follows.

Recall the parametrization of Pythagorean triples: $$a=2mn,\quad b=m^2-n^2,\quad c=m^2+n^2.$$

Compare this with the equation

$$10^{2k}x^2+y^2=z^2$$ that the OP derived.

We want to set things up in such a way that $a=10^kx$, $b=y$ and $c=z$. A way to use this is to select $m=m'\cdot 5^k$ and $n=n'\cdot 2^k$, and then arrange $y=m^2-n^2$, or actually $y^2$, to have the correct number of digits. This is not entirely trivial, but probably can be analyzed further without having to work too hard.

Example. The choice $m=2\cdot 5^4, n=78\cdot 2^4$ gives the following $$ \begin{aligned} a&=3\,120\,000,\\ x&=312,\\ y&=4996,\\ z&=3\,120\,004,\\ x^2&=97\,344,\\ y^2&=24\,960\,016,\\ z^2&=9\,734\,424\,960\,016. \end{aligned} $$ Observe that taking out the common factor $2$ from these choices of $m$ and $n$, IOW $m=5^4$, $n=39\cdot2^4$, leads to the following problem $$ \begin{aligned} a&=780\,000,\\ x&=78,\\ y&=1249,\\ z&=780\,001,\\ x^2&=6084,\\ y^2&=1\,560\,001,\\ z^2&=608\,401\,560\,001.\\ \end{aligned} $$ Undoubtedly you spotted the unwanted extra zero in the decimal expansion of $z^2$.

A smaller example. With $a=5^2$, $b=6\cdot2^2$ we get $$ \begin{aligned} a&=1200,\\ x&=12,\\ y&=49,\\ z&=1201,\\ x^2&=144,\\ y^2&=2401,\\ z^2&=1\,442\,401.\\ \end{aligned} $$

More examples. $$ \begin{array}{c|c|c|c|c|c} m&n&x&y&z&z^2\\ \hline 5^5&95\cdot2^5&19&524025&19007225&361274602200625\\ 10\cdot5^6&2441\cdot2^6&4882&8124324&48820000676&2383392466004640456976\\ 3\cdot5^7&1830\cdot2^7&1098&63263025&109800018225&12056044002210332150625 \end{array} $$

  • $\begingroup$ Alas, I'm no longer sure that it would be trivial to show that we can proceed indefinitely. $\endgroup$ – Jyrki Lahtonen Jul 5 '17 at 12:52

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