# When is a number like "ddd...ddd"+1 (where d is a digit) a perfect square or a prime?

Inspired by Is the number $333{,}333{,}333{,}333{,}333{,}333{,}333{,}333{,}334$ a perfect square?, I wonder when numbers like these are perfect squares. Certainly, all numbers of the form $$000...0001$$ are equal to the (non-prime) square $$1$$, and numbers of the form $$999...9990$$ are never primes (although sometimes they're $$90$$ times a repunit prime) and never squares (unless $$0$$ counts) because $$90$$ is divisible by $$2$$ and not $$4$$.

Therefore, the question I have boils down to:

### For which integers $$d\in[1,8]$$ and $$n\ge1$$ is $$1+d*\sum_{j=0}^n10^j=1+d*\left(10^{n+1}-1\right)/9$$ a perfect square? a prime?

Here are some initial calculations and observations for each value of $$d$$:

1. Never prime and never a square because $$12$$ and $$112$$ are not squares and numbers ending in $$1112$$ are divisible by $$8$$ but not $$16$$. However, $$1112$$, $$1111112$$, and $$1111111111111112$$ are all $$8$$ times a prime.
2. $$23$$, $$223$$, $$22222223$$, $$22222222223$$, $$222222222222222222222222222222222223$$ are all prime.
3. Never prime and never a square because $$34$$ is divisible by $$2$$ and not $$4$$. However, $$34$$, $$334$$, $$3334$$, $$333334$$, $$3333333334$$, $$333333333334$$, $$333333333333334$$, and $$333333333333333333333333333333334$$ are all twice a prime.
4. Never prime and never a square because $$45$$ is divisible by $$5$$ and not $$25$$. Each one of these is $$5$$ times a $$d=8$$ number.
5. Never prime because $$56$$ is divisible by $$4$$. However, $$56$$, $$556$$, $$555556$$, and $$555555555555556$$ are all four times a prime.
6. $$67$$, $$666667$$, $$66666667$$, $$666666667$$, $$66666666667$$, $$66666666666666666667$$, and $$66666666666666666666667$$ are all prime.
7. Never prime and never a square because $$78$$ is divisible by $$2$$ and not $$4$$. However, $$78$$, $$778$$, $$7777778$$, $$777777777777777778$$, and $$777777777777777777778$$ are all twice a prime.
8. $$89$$, $$88888888888889$$, $$88888888888888889$$, and $$88888888888888888888888888888888889$$ are all prime.

I don't know how to even approach "are there infinitely many primes for some $$d$$?" (and that may be hard since it's not known if there are infinitely many repunit primes) or proving that there are no squares (but I have more hope that that's solvable).

• All primes other than 2 and 3 are $6n+1$ or $6n-1$ form. Nov 2, 2013 at 1:27
• @mj9973 Good point. That at least narrows down which $d=2$ and $d=8$ numbers might be prime. Nov 2, 2013 at 1:31
• Also $99999\dots 99999+1=10000000\dots 00000$ and this is always square for even sets of $9$'s... Nov 2, 2013 at 1:41
• @abiessu But of course. I didn't mention that because 100...00 doesn't have a last digit different from its penultimate digit; it looks different. Nov 2, 2013 at 1:49
• Usually , $0 \cdots 01$ is not considered as a number because it has leading zeros. Apr 1, 2018 at 9:42

This answer proves that there are no such squares.

Proof :

Suppose that there are positive integers $$d,n,k$$ such that $$d\leqslant 8$$ and $$1+d\cdot\frac{10^{n+1}-1}{9}=k^2$$ i.e. $$9+d\cdot 10^{n+1}-d=9k^2$$ from which we have $$1-d\equiv k^2\pmod 4$$

• You already know that $$d$$ cannot be $$1,3,4,7$$.

• $$d$$ cannot be $$2,6$$ because supposing that $$d=2,6$$ gives $$k^2\equiv 3\pmod 4$$ which is impossible.

• For $$d=5$$, we have $$2^{n+1}\cdot 5^{n+2}=(3k-2)(3k+2)$$, so there have to be non-negative integers $$p,q,r,s$$ such that $$q+s\geqslant 3,3k-2=2^p5^q$$ and $$3k+2=2^r5^s$$ from which $$4=2^r5^s-2^p5^q$$ follows.

• For $$d=8$$, we have $$2^{n+4}\cdot 5^{n+1}=(3k-1)(3k+1)$$, so there have to be non-negative integers $$p,q,r,s$$ such that $$q+s\geqslant 2,3k-1=2^p5^q$$ and $$3k+1=2^r5^s$$ from which $$4=2^{r+1}5^s-2^{p+1}5^q$$ follows.

So, in order to prove that $$d$$ cannot be $$5,8$$, it is sufficient to prove that there are no non-negative integers $$p,q,r,s$$ such that $$q+s\geqslant 2$$ and $$4=2^r5^s-2^p5^q$$.

If $$q\geqslant 1$$ and $$s\geqslant 1$$, then RHS is divisible by $$5$$ while LHS isn't. So, we have either $$q=0$$ or $$s=0$$.

(Case 1) For $$q=0$$, we have $$4=2^r5^s-2^p$$. If $$p=0$$, then $$5=2^r5^s$$ implies $$(p,q,r,s)=(0,0,0,1)$$ which does not satisfy $$q+s\geqslant 2$$. If $$p=1$$, $$6=2^r5^s$$ is impossible. So, $$p\geqslant 2$$ and then since $$r\geqslant 2$$, we have $$1=2^{r-2}5^s-2^{p-2}$$. If $$r-2\geqslant 1$$ and $$p-2\geqslant 1$$, then RHS is divisible by $$2$$ while LHS isn't. So, we have either $$r-2=0$$ or $$p-2=0$$. If $$p-2=0$$, then $$2=2^{r-2}5^s$$ so $$(p,q,r,s)=(2,0,3,0)$$ which does not satisfy $$q+s\geqslant 2$$. If $$r-2=0$$, then $$1=5^s-2^{p-2}$$. Since $$1\equiv (-1)^s-(-1)^{p-2}\pmod 3$$, $$s$$ has to be odd. So, letting $$s=2t+1$$, we have $$2^{p-2}=5\cdot 25^t-1$$. If $$p-2\geqslant 3$$, then LHS is divisible by $$8$$ while RHS isn't. So, we have $$p-2\lt 3$$, and so $$(p,q,r,s)=(4,0,2,1)$$ which does not satisfy $$q+s\geqslant 2$$.

(Case 2) For $$s=0$$, we have $$4=2^r-2^p5^q$$. If $$r=0$$, then $$3=-2^p5^q$$ is impossible. If $$r=1$$, then $$2=-2^p5^q$$ is impossible. So, $$r\geqslant 2$$, and then since $$p\geqslant 2$$, we have $$1=2^{r-2}-2^{p-2}5^q$$. If $$r-2\geqslant 1$$ and $$p-2\geqslant 1$$, then RHS is divisible by $$2$$ while LHS isn't. So, we have either $$r-2=0$$ or $$p-2=0$$. If $$r-2=0$$, then $$0=-2^{p-2}5^q$$ is impossible. So, $$p-2=0$$ gives $$1+5^q=2^{r-2}$$. If $$r-2\geqslant 2$$, then RHS is divisible by $$4$$ while LHS isn't. So, we have $$r-2\lt 2$$, and so $$(p,q,r,s)=(2,0,3,0)$$ which does not satisfy $$q+s\geqslant 2$$.

In conclusion, there are no such squares.$$\quad\blacksquare$$