Soumya M. answered 03/10/24
I am a PhD student who is very passionate about STEM education.
Hi Evelyn, this question looks interesting and I have come up with a solution. I apologize for the text. I'm new here and this editor does not seem to support latex. Make sure you write the proof along as you read because it will be really helpful.
Let α= $\sqrt{16n+21}$ and
let β= $\sqrt{16n+24}$.
We first have to prove that α and β can't be integers individually. Let us start with alpha first. We will use the method of proof by contradiction. We can write,
α2= $16n+21$.
Now note that if we assume that there exists an integer α such that the above equation holds, then it must be an odd integer.
So, let us assume α=$2y+1$. Now, substituting this form of α in the above equation we get
$(2y+1)^2=16n+21$
Expanding,
$4y^2+4y+1=16n+21$
or,
$4y(y+1)=4 {4(n+1)+1}$
simplifying further and assuming, k=(n+1) we get,
$y^2+y=4k+1$.
Now the interesting thing here is that the RHS i.e, 4k+1=1 mod 4. But the LHS i.e, y^2+y= 0 or 2 mod 4. So clearly our assumption of α being integer was incorrect.
Now, let us focus on β. Taking a factor of 4 common throughout we can write,
β= $\sqrt{4[4(n+1)+2]}=2\sqrt{[4(n+1)+2]}$.
Again, writing n+1=k. We have
β=$2\sqrt{[4k+2]}$.
Note again, that 4q+2 or the quantity under square root is never a perfect square. Because, a perfect square has to be 0 or 1 mod 4. But, 4q+2= 2 mod 4.
So far we have proved that α and β can never be integers. Now lets prove that no integer can lie between them. For that let us subtract square of β from square of α.
We get, α2-β2=3.
Now,
(β+1)2-β2= 2β+1.
This implies that for β>1 we always have
(β+1)2-β2= 2β+1 >α2-β2= 3.
Or,
(β+1)2 > α2.
Since, everything is positive, we have
α< β+1.
Or,
α-β < 1.
So no integer lies between them. Hence proved.
If you liked the proof or have any questions please comment. I'm always there to help :)

Paul M.
03/11/24
Soumya M.
Paul M. I see what you’re saying. The condition I have proved is not strong enough. Btw I don’t see any problem with algebra. If my idea is correct I should have a right solution but it is too tedious to write it down. I’ll do it tomorrow.03/13/24

John C.
I agree with Soumya, the solution looks good! She'd only have to check 16n + 22 and 16n+23 also aren't squares (which she's basically already done with the mod 4 trick) since any integer squared in that range would have to be one of the 4 (16n+1 thru 16n+4). I also tried doing this problem, but with difference of squares and ran into trouble, so it's really nice to see a proof using modular arithmetic! Wasn't sure what background the poster assumed was necessary for the problem.03/16/24
Evelyn D.
Thank you so much!03/11/24