Under what conditions is it true that $m^2 - a$ is not a square if and only if $(m - 1)^2 < m^2 - a < m^2$, where $a>0$?

Question

Preamble: The present inquiry is an offshoot of this MSE question from February 21, 2019, and ultimately, Theorem III.2, page 2 from a paper submitted to a conference organized by De La Salle University-Manila. I was supposed to deliver a talk regarding this topic on September 22, 2009, but I was unable to attend, due to a schedule conflict with my work back then. (My apologies.) The main impetus for this latest MSE question is my having recently found this MSE question.

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PROBLEM STATEMENT

Under what conditions is it true that $$m^2 - a \text{ is not a square } \iff (m - 1)^2 < m^2 - a < m^2,$$ where $a>0$?

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MY ATTEMPT

Let $a>0$. If $m^2 - a$ satisfies the inequality $$(m - 1)^2 < m^2 - a < m^2$$ then $m^2 - a$ is obviously not a square, as it is between two consecutive squares.

Alas, this is where I get stuck, as I currently do not know how to prove the converse.

SEARCH FOR COUNTEREXAMPLES

I ran the following Pari-GP script to search for counterexamples to the converse, however, no output was returned in the range $1 \leq m \leq {10}^{10}$ and $1 \leq a \leq {10}^{10}$:

for(x=1, 10000000000, for(y=1, 10000000000, if(!issquare(x^2 - y) && (x^2 - y <= (x-1)^2),print(x,"     ",factor(x),"     ",y,"     ",factor(y)))))

OPTIMIZED SCRIPT (communicated by Brian Moehring):

for(x=2, 105, for(y=2*x, x^2, if(!issquare(x^2 - y),print(x,"     ",y))))

I am therefore led to believe that the criterion indeed holds. (There might have been a bug/error with Pari-GP script processing in Sage Cell Server while I was executing my scripts. I am not an expert on determining computer time/space requirements for running a program/script, but Brian Moehring has already pointed out in this comment that my original script would have caused the servers to fail to give an output for my computational request, so I will have to defer to his opinion.) As pointed out by TonyK in a comment, $m=5$ and $a=15$ is a counterexample to the converse. (There are literally a lot of them.)

I would, of course, still be interested in researching the conditions under which the criterion holds. Hence, this question.

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CONTEXT

Specifically, let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Then it is known that $m^2 - p^k$ is not a square, granted we could prove that $p < m$ and $m^2 - p^k$ is not a power of two.

UPDATE: Here is a conclusive proof that $m^2 - p^k$ is not a square. (It turns out that one does not need the extra assumption that $m^2 - p^k$ is not a power of two.) Assume to the contrary that $m^2 - p^k = n^2$. Then as proved in this MSE question, we derive $m = (p^k + 1)/2$. Now, in this answer, MSE user FredH proves that the assumption $k \neq 1$ contradicts $$\gcd(p^{k+1}-1, (p-1) p^k (p^k + 1)^2) \leq (p-1)\gcd(p^{k+1}-1,p^k+1)^2,$$ which follows from $$\sigma(p^k) \sigma(m^2) = 2p^k m^2 \tag{$**$}$$ and therefore, $$2(p^{k+1}-1) \sigma(m^2) = (p-1) p^k (p^k + 1)^2.\tag{$*$}$$ Equation $(*)$ is obtained by substituting $m = (p^k + 1)/2$ in Equation $(**)$. Since $m = (p^k + 1)/2 < p^k$, the assumption $p < m$ implies $k \neq 1$, whereupon we have the implication $$p < m \implies m^2 - p^k \text{ is not a square}.$$ By the contrapositive, we obtain $$m^2 - p^k \text{ is a square} \implies m < p \implies k = 1.$$ But then we also know that $$m^2 - p^k \text{ is a square} \iff m = (p^k + 1)/2 \iff p^k = 2m - 1.$$ But we already know that $k=1$ follows from the given assumption, so that $$p = 2m - 1.$$ However, Acquaah and Konyagin essentially proved in (IJNT, 2012) that if $p$ is the special prime factor of an odd perfect number $p^k m^2$, then the implication $$k = 1 \implies p < m\sqrt{3}$$ holds. Together with the implication $$k > 1 \implies p < m$$ as proved by Dris in (JIS, 2012), we know unconditionally that $$p < m\sqrt{3}$$ must hold. But recall that we have $2m - 1 = p$. We infer that $$2m - 1 = p < m\sqrt{3}$$ which is equivalent to $$m(2 - \sqrt{3}) < 1.$$ This contradicts the fact that $m$ is a humongous number. (In fact, one can compute the lower bound $m > {10}^{375}$, and this follows from Ochem and Rao's $N > {10}^{1500}$ (Math. Comp. (2012) and the estimate $p^k < m^2$ by Dris in (JIS, 2012).)

Hence, $m^2 - p^k$ is not a square. (Additionally, since $4 \mid (m^2 - p^k)$, then $m^2 - p^k$ is likewise not squarefree.)

Since $m^2 - p^k$ is not a square and $m^2 - p^k$ is divisible by $4$, then we infer that $m^2 - p^k = 2^r t$ where we necessarily have $r \geq 2$ and $2^r \neq t$. (I am not really sure whether the exponential Diophantine equation $m^2 - p^k = 2^r t$ will force $\gcd(2,t)=1$, if $r \geq 2$ and $2^r \neq t$.)

Notice that all of the variables $m, p, k, r,$ and $t$ are actually not known, a priori. But we may try testing heuristics (e.g. inequalities or equations) relating their values, using the Descartes spoof as an "experimental/combinatorial sample".

This particular topic is currently not my cup of tea. Hence, this question.

Answer 1

Under what conditions is it true that $$m^2 - a \text{ is not a square } \iff (m - 1)^2 < m^2 - a < m^2,$$ where $a>0$?

We know it is true that $$(m - 1)^2 < m^2 - a < m^2\implies m^2 - a \text{ is not a square }$$

So, what we want to know is that under what conditions it is true that $$m^2 - a \text{ is not a square } \implies (m - 1)^2 < m^2 - a < m^2\tag2$$

• A proof that $0\lt a\lt 2m-1$ is necessary : Suppose that $m^2-a$ is not a square with $a\geqslant 2m-1$. Then, we have $m^2-a\leqslant m^2-(2m-1)=(m-1)^2$. So, $(m-1)^2\lt m^2-a$ does not hold.

• A proof that $0\lt a\lt 2m-1$ is sufficient : If $m^2 - a$ is not a square with $0\lt a\lt 2m-1$, we have $m^2-a>m^2-(2m-1)=(m-1)^2$ and $m^2-a<m^2$.

Therefore, we can say that under the condition that $0\lt a\lt 2m-1$, it is true that $$m^2 - a \text{ is not a square } \iff (m - 1)^2 < m^2 - a < m^2$$

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From what we've got, we can say that under the condition that $p^k\lt 2m-1$, it is true that $$m^2 - p^k \text{ is not a square } \iff (m - 1)^2 < m^2 - p^k < m^2$$

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Since $m^2 - p^k$ is not a square and $m^2 - p^k$ is divisible by $4$, then we infer that $m^2 - p^k = 2^r t$ where we necessarily have $r \geq 2$ and $2^r \neq t$. (I am not really sure whether the exponential Diophantine equation $m^2 - p^k = 2^r t$ will force $\gcd(2,t)=1$, if $r \geq 2$ and $2^r \neq t$.)

For each positive integer $N$, if we define $u$ as a non-negative integer satisfying both $2^u\mid N$ and $2^{u+1}\not\mid N$, and $v$ as $\cfrac{N}{2^u}$, then there is only one such pair of integers $(u,v)$, and $u,v$ satisfy $$N=2^uv,\qquad u\geqslant 0,\qquad v\gt 0,\qquad v\equiv 1\pmod 2$$

Examples :

• For $N=24$, $(u,v)=(3,3)$.

• For $N=59$, $(u,v)=(0,59)$.

• For $N=138$, $(u,v)=(1,69)$.

• For $N=1024$, $(u,v)=(10,1)$.

Similarly, if we define $r$ as a non-negative integer satisfying both $2^r\mid m^2-p^k$ and $2^{r+1}\not\mid m^2-p^k$, and $t$ as $\cfrac{m^2-p^k}{2^r}$, then there is only one such pair of integers $(r,t)$, and $r,t$ satisfy $$m^2-p^k=2^rt,\qquad r\geqslant 2,\qquad t\gt 0,\qquad t\equiv 1\pmod 2$$

So, if you define $r,t$ as above, then the integers $r,t$ are not independent from $m,p$ and $k$. The integers $r,t$ are determined by $m,p$ and $k$, so the $r$ means $r(m,p,k)$, and the $t$ means $t(m,p,k)$. Since $t$ is, by the definition, odd, we can say that $\gcd(2,t)=1$.

If you don't define $r,t$ as above (in other words, if you are just writing $m^2-p^k=2^rt,r\geqslant 2,2^r\not=t$), then there are many such pairs $(r,t)$. For example, if $m^2-p^k=192$, then we have $192=2^2\cdot 48=2^3\cdot 24=2^4\cdot 12=2^5\cdot 6=2^6\cdot 3$, so $(r,t)=(2,48),(3,24),(4,12),(5,6),(6,3)$. So, $t$ can be even. On the other hand, if you define $r,t$ as above, we get only one pair $(r,t)=(6,3)$ where $t$ is odd.