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So I was thinking of a particular problem the other day and started experimenting with it. It goes like this:

I want to find natural numbers $x$ and $y$ such that:

$x$ is an $n$-digit number, and $y$ is an $m$-digit number, where $n$ and $m$ are not necessarily equal. Let their digit representations be $x = a_1a_2...a_n$ and $y = b_1b_2...b_m$, where all $a_i$ and $b_j$ are between $0$ and $9$, except for $a_1$ and $b_1$. Neither $a_1$ nor $b_1$ can be equal to $0$.

Also, let $x'$ have the digit representation $a_{1}a_{2}...a_{n} 2$, and $y'$ have $b_1b_2...b_m2$. I call $x'$ and $y'$ the "lazy squares" (hehe) of $x$ and $y$ if the following holds true: $$x^2 + y^2 = x' + y'.$$

In general this seems tedious and hard. What I have done thus far is trying a simpler case, supposing that both $x$ and $y$ are $1$-digit numbers. Now their only digits are simply $x$ and $y$ respectively. This yields the following equation: $$x^2+y^2 = 10x+10y+4 .$$ After some algebraic manipulations we can rewrite the equation as the equation of a circle in the following way: $$(x-5)^2+(y-5)^2 = 54.$$

Now, I came to the conclusion that this equation does not have any natural numbered solutions just by eye-balling it in desmos. I guess you could do an exhaustive solution/proof by making a program in python for instance and just check for all $1$-digit $x$'s and $y$'s and come to the same conclusion.

So... I am wondering if there might be a simpler solution or approach to the general problem that I'm not able to see? If not, I'm wondering how one could take a stab att the more general case where $x$ and $y$ are $n$-digit and $m$-digit natural numbers respectively?

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Victor Galeano
  • 559

1 Answers1

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As @Jyrki Lahtonen commented the formula $$(x-5)^2+(y-5)^2 = 54$$ actually holds for any $n$-digit number $x$ and $m$-digit number $y$. I will rewrite this slightly with some change of variables: $u=x-5$ and $v=y-5$ yielding $$u^2+v^2 = 54.$$ From this answer: https://math.stackexchange.com/a/124173/682544 I saw that one could count the number of lattice points on the circle (AKA the number of integer coordinates on the circle) by calculating:

  1. $C_1:=$ Number of divisors $d$ of $54$ such that $d\equiv 1\pmod {54}$
  2. $C_3:=$ Number of divisors $d$ of $54$ such that $d\equiv 3\pmod {54}$
  3. The number of integer solutions then is equal to $4(C_1-C_3)$ (apparently it has something to do with gaussian integers and primes).

Now, the divisors of 54 are: $1$, $2$, $3$, $6$, $9$, $18$, $27$ and $54$ itself. So $C_1=2$ and $C_3=2$. Thus the equation under consideration has $4(2-2)=0$ integer solutions for $u$ and $v$ which in turn implies that $x$ and $y$ can not be integers.

Is this correct? It anoys me that I dont understand why one can use the above algorithm. I tried watching this video to understand it all: https://www.youtube.com/watch?v=rr9l9q22r9o but for the moment I am not convinced. Hehe.

Victor Galeano
  • 559