Find all positive integers x, y, z so that:
 x^{2} + y = z and:
 Base ten expansion of each of x and y has precisely n digits and that of z has precisely 2n digits, where n > 1.
 All the digits of x are the same, all the digits of y are the same and all the digits of z are the same.
Prove that there are no others.
For n=1 see Charlie's solution.
This is a proof for n>1
Let a be the repeated digit of x so x=a(10^n  1)/9
Let b be the repeated digit of y so y=b(10^n  1)/9
Let c be the repeated digit of z so z=c(10^(2n)  1)/9
so x^2 + y = z becomes
a^2[10^(2n)  2*10^n + 1]/81 + b(10^n  1)/9 = c(10^(2n)  1)/9
For values of a that are not multiples of 3, there won't be an integer value of c that balances the 10^(2n) term
If a=3 we get
[10^(2n)  2*10^n + 1] + b(10^n  1) = c(10^(2n)  1)
to balance the 10^(2n) term, c must be 1
and it then follows that b=2 balances the rest.
If a=6 we get
4*[10^(2n)  2*10^n + 1] + b(10^n  1) = c(10^(2n)  1)
to balance the 10^(2n) term, c must be 4
and it then follows that b=8 balances the rest.
If a=9 the c would have to be 3 but the b that balances the rest would be 18. (The value of y would have overlapping 18's and look more like 199...998.)

Posted by Jer
on 20140512 12:47:31 