Quadratic residues that are also squares themselves

Question

I have a question about number bases and square numbers. When considering quadratic residues in a given modulus (without eliminating residues that are not relatively prime to the modulus), it is evident that in a number of smaller moduli the squares that are less than the modulus comprise all the quadratic residues that it has, such that there are not any "false squares." This is the same question as asking whether a square when written in a given number base can only have a square in the one's place. For example, in base ten this is not the case, since a square can end in 0,1,4,5,6,9, and 5 and 6 are not squares. On the other hand, in both the duodecimal and hexadecimal systems a square can only end in 0,1,4,9, all of which are squares in themselves. For which number bases is this always the case, and is there a standard proof for it?

Answer 1

the observations on small numbers do lead to a proof. First, the number $d(n)$ of positive divisors of $n$ satisfies $d(n) < 2 \sqrt n,$ simply because each divisor $u > \sqrt n$ pairs up with a divisor $\frac{n}{u} < \sqrt n.$ Even if $n$ itself is a square, you get just one extra.... We may need a stronger one, $$ d(n) \leq 48 \sqrt[3]{\frac{n}{2520}} $$ with equality only at $n= 2520.$

Now, suppose $B \geq 17.$ For every number $x$ with $\sqrt B < x < \sqrt{2B}, $ we see $x^2$ lies between $B$ and $2B.$ Furthermore, the reduction $\pmod B$ is then exactly $x^2 - B .$ How often can this be a square? If so, we have some $y < \sqrt B$ such that $$ x^2 - B = y^2 $$ or $$ x^2 - y^2 = (x+y)(x-y) = B $$ Every time we find such an $x,y$ we have a (disticnt) divisor $(x+y),$ one for each value of $x$ that works out. The count of $x$ values is $\sqrt{2B} - \sqrt B \approx \frac {\sqrt B}{\sqrt 2 + 1}.$

For large enough $B$ this is much more than $48 \sqrt[3]{\frac{B}{2520}} $ and there are $x$ values that do not lead to divisors of $B;$ such an $x$ means failure.

Enough for now.

just numerical illustration, I ran all $B \leq 1000000. $ The bound with the cube root says $d(1000000) < 352.73 \; \; .$ In comparison, $\sqrt{2000000} - \sqrt{1000000} \approx 414.21 \; \; , $ which is already bigger than the cube root thing.