# Prove that the product of any two numbers between two consecutive squares is never a perfect square

In essence, I want to prove that the product of any two distinct elements in the set $$\{n^2, n^2+1, ... , (n+1)^2-1\}$$ is never a perfect square for a positive integers $$n$$. I have no idea on how to prove it, but I've also yet to find a counterexample to this statement. Can anyone help?

• just inequalities. All positive integers, can we have $n^2 < 2 w^2 < 2 (w+1)^2 < (n+1)^2 \; \; ? \; \;$ How about $n^2 < 3 w^2 < 3 (w+1)^2 < (n+1)^2 \; \; ? \; \;$ – Will Jagy Feb 17 at 20:17
• is $n^2$ included in the set? If so, why is $(n+1)^2$ not included? – Dr. Mathva Feb 17 at 20:17
• @Dr.Mathva Because then the statement is obviously false, with (n(n+1))^2 – gnasher729 Feb 17 at 20:20

## 2 Answers

First, note that $$n^2$$ can't be one of the elements as the other element would also need to be a perfect square for the product to be a perfect square. As gnasher729 commented to the answer, this is why $$\left(n + 1\right)^2$$ is not included.

Assume there are $$2$$ such elements, $$n^2 + a$$ and $$n^2 + b$$, with $$a \neq b$$, $$1 \le a, b \le 2n$$ and, WLOG, $$a \lt b$$. Thus, consider their product to be a perfect square of $$n^2 + c$$, i.e.,

$$\left(n^2 + a\right)\left(n^2 + b\right) = \left(n^2 + c\right)^2 \tag{1}\label{eq1}$$

for some integer $$a \lt c \lt b$$. As such, for some positive integers $$d$$ and $$e$$, we have that

$$a = c - d \tag{2}\label{eq2}$$ $$b = c + e \tag{3}\label{eq3}$$

Substitute \eqref{eq2} and \eqref{eq3} into \eqref{eq1} to get

$$\left(n^2 + \left(c - d\right)\right)\left(n^2 + \left(c + e\right)\right) = \left(n^2 + c\right)^2 \tag{4}\label{eq4}$$

Expanding both sides gives

$$n^4 + 2cn^2 + \left(e - d\right)n^2 + c^2 + c\left(e - d\right) - ed = n^4 + 2cn^2 + c^2 \tag{5}\label{eq5}$$

Removing the common terms on both sides and moving the remaining terms, apart from the $$n^2$$ one, to the right gives

$$\left(e - d\right)n^2 = -c\left(e - d\right) + ed \tag{6}\label{eq6}$$

Note that $$e \le d$$ doesn't work because the LHS becomes non-positive but the RHS becomes positive. Thus, consider $$e \gt d$$, i.e., let

$$e = d + m, \text{ where } m \ge 1 \tag{7}\label{eq7}$$

Using \eqref{eq3} - \eqref{eq2}, this gives

$$b - a = e + d \lt 2n \Rightarrow 2d + m \lt 2n \Rightarrow d \lt n - \frac{m}{2} \tag{8}\label{eq8}$$

Also,

$$ed = \left(d + m\right)d \lt \left(n + \frac{m}{2}\right)\left(n - \frac{m}{2}\right) = n^2 - \frac{m^2}{4} \lt n^2 \tag{9}\label{eq9}$$

Since $$e \gt d$$ means that $$-c\left(e - d\right) \lt 0$$, the RHS of \eqref{eq6} cannot be a positive integral multiple of $$n^2$$, so it can't be equal to the LHS.

• Or, by (7), and by $c>0,$ since $e-d$ and $c$ are positive we have, in (6) that $n^2\le (e-d)n^2<ed$ which contradicts (9)....+1. – DanielWainfleet Feb 18 at 21:49

For any two numbers $$n^2+a,\ n^2+b;\ 0, their product will satisfy $$n^4.

All of the squares between $$n^4$$ and $$(n^2+2n+1)^2$$ will have the form $$(n^2+m)^2=n^4+2mn^2+m^2;\ 1\le m\le 2n$$

If $$n^4+(a+b)n^2+ab$$ is a perfect square, it will be one of the squares between $$n^4$$ and $$(n^2+2n+1)^2$$ and hence equal to $$n^4+2mn^2+m^2$$ for some $$m$$.

Thus $$(a+b)=2m;\ ab=m^2$$ for some $$m$$

Rearranging, we get $$m^2=\frac{a^2+2ab+b^2}{4}=ab=\frac{4ab}{4}$$, or $$a^2+b^2=2ab$$.

This implies both $$a\mid b$$ and $$b\mid a$$, meaning $$b=a$$ and the numbers being multiplied to obtain a perfect square are not distinct.

Added by edit: John Omielan comments (for the specific case $$k=1$$) that my original answer fails to consider possible solutions of the form $$a+b=2m-k;\ ab=m^2+kn^2$$. He separately provides a more complete answer that addresses those cases. Marty Cohen comments that I can only properly conclude $$n^4+(a+b)n^2+ab=n^4+2mn^2+m^2 \Rightarrow n^2(a+b-2m)=(m^2-ab)$$. Let me address those shortcomings.

If $$(n^2+a)(n^2+b)=(n^2+m)(n^2+m)$$ then either $$a=b=m$$ (addressed in my original answer) or $$a (which this edit will address). If $$n^2(a+b-2m)=(m^2-ab)$$, then $$(m^2-ab)$$ is divisible by $$n^2$$, so $$(m^2-ab)=kn^2=n^2(a+b-2m)$$, or $$(a+b-2m)=k\Rightarrow a+b=2m+k$$.

$$m,a,b$$ have limits on their sizes. $$m. Also $$a Hence, $$|(m^2-ab)|<4n^2 \Rightarrow |k|=(a+b-2m)<4$$. For $$(a+b)=2m+k,\ |k|=0,1,2,3$$, where $$k=0$$ corresponds to the case where $$a=b=m$$.

The midpoint $$t$$ between $$a$$ and $$b$$ is the average $$t=\frac{a+b}{2}$$. Let $$b-t=r,\ t-a=r$$. Note that if $$a$$ and $$b$$ have different parity, $$t$$ and $$r$$ may have half integral values. Finally, $$b-a=2r$$, but since $$b\le2n,\ a\ge 1$$, then $$2r=b-a<2n\Rightarrow r.

$$(n^2+a)(n^2+b)=((n^2+t)-r)((n^2+t)+r)=(n^2+t)^2-r^2$$. Therefore, unless $$m>t$$, $$(n^2+a)(n^2+b)<(n^2+m)^2$$. But $$t=\frac{a+b}{2}=m-\frac{k}{2}$$. $$m-t=\frac{1}{2},1,\frac{3}{2}$$. Within the constraints of the original question, $$m$$ must be larger than the average of $$a$$ and $$b$$, but it must be very close to that average.

Substituting $$t-\frac{k}{2}$$ for $$m$$ in $$(n^2+m)^2$$, we get $$((n^2+t)-\frac{k}{2})^2$$. Can that equal $$(n^2+a)(n^2+b)=((n^2+t)-r)((n^2+t)+r)=(n^2+t)^2-r^2$$? Letting $$s=(n^2+t)$$ for simplicity in keeping track during expansion, we ask whether $$(s-\frac{k}{2})^2=(s-r)(s+r)=s^2-r^2\Rightarrow ks-\frac{k^2}{4}=r^2\Rightarrow k(n^2+t)-\frac{k^2}{4}=r^2$$?

$$r. It is obvious for $$3\ge k>1$$ that $$k(n^2+t)-\frac{k^2}{4}>n^2>r^2$$. For $$k=1$$, $$n^2+t-\frac{1}{4}>n^2 \iff t>\frac{1}{4}$$. For $$a, $$\min(a+b)=4 \Rightarrow \min(t)=2; 2>\frac{1}{4}$$.

There are no solutions to the question for $$a.

• Simpler: if $a+2+b^2 = 2ab$ then $(a-b)^2 = 0$ so $a = b$. Nice answer. – marty cohen Feb 17 at 21:48
• Note that $n$ is a fixed value. How do you know that, for example, $a + b = 2m - 1$ and $ab = n^2 + m^2$ can't be true instead? It might be intuitively obvious to you, & others, but it's not to me. This is why I try to explicitly show this doesn't occur in my answer. – John Omielan Feb 17 at 21:49
• I think this answer is wrong. All you can conclude is that $(a+b)n^2+ab = 2mn^2+m^2$ or $n^2(a+b-2m) = m^2-ab$. This is not a polynomial identity in $n$, so you can not conclude that $a+b = 2m$ and ab = n^2\$. – marty cohen Feb 18 at 17:59