Determine all the possible positive integer $n$,such that $3^n+n^2+2019$ is a perfect square.

Question

Problem: Determine all the possible positive integer $n$,such that $3^n+n^2+2019$ is a perfect square.

My approach: At first I used guessing and checking. By that I found $n=4$. Then, I tried more values of $n$ and got no result. Then, I was pretty sure that there is no solution for $n>4$. But, I have to prove it. Then, I observed that $n$ has to be even. then, I tried to prove there is no solution for $n>4$. I used induction but failed.

Conclusion: How to solve this? It's a math contest practice problem. I am preparing for math contest. I was quite upset because after $2$ years of training, I can't solve it. Is this question junior standard(grade $6$ to $8$)? If yes then will it be eligible for regional round or national round? Any suggestion is appreciated.

Hint: If $n=2m$ then note that $(3^m+1)^2$ grows very rapidly in $m$. If you can show that it is larger than your expression, you are done. This is a standard way to show that a natural number isn't a square...argue that it is between two consecutive perfect squares. — lulu, Mar 06 '25 at 13:46
should add: questions concerning the make up of some upcoming competition or exam are not appropriate for this site. We don't know how the contest(s) will be set up. Ask your instructors or coaches. — lulu, Mar 06 '25 at 13:47
For $n\ge5 $ , $3^n$ grows exponentially compared to $n^2 + 2019$ — Ved Patel, Mar 06 '25 at 13:52
@lulu How can I show something grows rapidly? I don't know that well. Also, I tried a lot. — Geometry99, Mar 06 '25 at 14:00
To repeat my hint: you are done if you can show that $(3^m+1)^2>3^{2m}+4m^2+2019$. Of course, this isn't true for all $m$, but it is certainly true for most $m$. — lulu, Mar 06 '25 at 14:02
$3^n+n^2+2019 = x^2 \iff 3^n+2019 = x^2 - n^2 \iff 3 \cdot (3^{(n-1)} + 673) = (x-n) \cdot (x+n)$, ..., does this help? — Dominique, Mar 06 '25 at 14:02
@lulu Why would I be done if I can show $(3^m+1)^2>3^{2m}+4m^2+2019$? What's the reason? — Geometry99, Mar 06 '25 at 14:09
As I said in the hint: If your expression is $f(n)$, with $n=2m$ , then proving that inequality would show that $(3^m)^2<f(n)<(3^m+1)^2$ so that $f(n)$ would lie between two consecutive perfect squares. — lulu, Mar 06 '25 at 14:13
FYI, using an Approach0 search, there's the AoPS thread 3^n+n^2+2019 is a perfect square, which gives the problem source as being "$2019$ China Western Mathematical Olympiad P$1$". This gives several solutions, so once you understand one of them, you may wish to reference it and post your own answer here based on it. — John Omielan, Mar 06 '25 at 15:32
See this similar question where the answer opens with a strategy you can apply here. — 2'5 9'2, Mar 06 '25 at 18:42

score 1 · Answer 1 · answered Mar 14 '25 at 14:32

We define $f: \mathbb{N}\to \mathbb{N}$ as:

$f(n) =3^n+n^2+2019$

Lemma 1. if $x$ is square then $x \equiv 1 $ (mod 4) or $x \equiv 0$ (mod4).

Proof. $r = 2k \Rightarrow r^2\equiv 0$(mod 4) $r = 2k+1 \Rightarrow r^2\equiv 1$(mod 4) .$\square$

Proposition 1. if $f(n)$ is square then $ n = 2k $.

Proof.$f(2k+1) = 3^{2k+1}+(2k+1)^2+2019\equiv 3+1+3 \equiv 3 $ (mod 4) $f(2k) = 3^{2k}+(2k)^2+2019\equiv 1+0+3 \equiv 0 $ (mod 4).$\square$

Theorem 1 . For $k\ge 4$ then $3^{2k}< f(2k) < (3^k +1)^2$

Proof. it is clear that $ 3^{2k} < 3^{2k} + (2k)^2 + 2019 $ now we proof $f(2k) < (3^k +1)^2$ by induction:

$$ k =4 : f(2k) = 3^{2k}+(2k)^2+2019 < 3^{2k} + 2\cdot 3^{k} +1 \Leftrightarrow (2k)^2+2019< 2\cdot 3^{k} +1 $$

$$(2k)^2+2019 = 8^2 +2019 < 80\cdot 80 = 6400 < 81^2 + 1< 2\cdot 3^{k} +1 $$

so for $k = 4 $ it is correct now assume for $k $ it's true for $k+1 $ we have:

$$ f(2k+2) < 3^{2k} + 2\cdot 3^{k} +1 \Rightarrow (3^{k+2} +1)^2 > 9(3^{2k} + 2\cdot 3^{k} )+ 1 > 9(3^{2k}+(2k)^2+2018)+1 $$

$$ \Leftrightarrow (3^{k+2} +1)^2> 3^{2(k+1) } +9 (2k)^2+2019 $$

and it is obvious that $ 9 (2k)^2 = 36k^2>16k^2 = 4k^2 + 8k^2 + 4k^2 > 4k^2 + 4k+4 = (2k+2) ^2$

So we got:

$$ (3^{k+2} +1)^2 >3^{2(k+1) } + (2k+2)^2+2019$$.$\square$

Know we should check $n\in \{2,4,6\}$ because by Theorem 1 we proof that if $n \in \{8,10,12 ,\cdots\}$ then $f(n)$ is between two consecutive squares so it can't be square now we have:

$n =2 : 45^2 = 2025< f(2) = 2032 < 2116 = 46^2$

$n= 4 : f(4) = 2116= 46^2$ it is square

$n =6 : 45^2 = 52^2 = 2704< f(2) = 2784 < 2809 = 53^2$

So just if $n =4$ then $f(n) $ is square!

Hoin Ryu · Answer 2 · 2025-03-09T13:28:48.350

$3^n$ is much more larger than $n^2$ if n is large enough, so we can think that if $3^n$ is square then $3^n+n^2+1$ won't be square since it will be less then $(3^{n/2}+1)^2$. Let's say $m^2=3^n+n^2+2019$, and suppose that n is odd. Then $m^2 \equiv -1+1+3 \equiv 3 (mod 4)$, and this is a contradiction. Thus n is even, and we can say n=2k. Since $(3^k)^2 < m^2 < (3^k+1)^2=3^n+2\times3^k+1$ if $k \geq 7$, so k must be less than 7, and we have to check all cases to find n.

k=6 : $m^2=3^{12}+2163 \Rightarrow (3^6+1)^2<m^2<(3^6+2)^2$
k=5 : $m^2=3^{10}+2119 \Rightarrow (3^5+4)^2<m^2<(3^5+5)^2$
k=4 : $m^2=8644, 92^2=8464, 93^2=8649$
k=3 : $m^2=2784, 52^2=2704, 53^2=2809$
k=2 : $m^2=2116=46^2$
k=1 : $m^2=2032, 45^2=2025, 46^2=2116$

Finally,the solution is n=2.

Determine all the possible positive integer $n$,such that $3^n+n^2+2019$ is a perfect square.

2 Answers2