Can the given expression equal a perfect square?

“I fear not the man who has practised 10,000 kicks once, but I fear the man who has practised one kick 10,000 times.”

- Bruce Lee

Certain classes of problem-solving strategies repeatedly prove helpful in many Olympiads worldwide. In today’s blog post, we will discuss some strategies for the following type of problem:

Find all integers $n$ such that a quadratic expression $f(n)$ (or $f(n,m,\ldots)$) is a perfect square.

There are usually two strategies that are used to claim no solutions exist:

• Using quadratic residues, argue that there are no solutions. For instance, $12m+ 19$ can never be a perfect square since it is for the form $4k+3$ and we know that 3 is not a quadratic residue modulo 4.
• Bound the given expressions between two consecutive perfect squares. For instance, $n^2 +n+1$ can be bounded as follows: \[n^2 < n^2+n + 1< (n+1)^2.\] But there cannot be a perfect square between two consecutive perfect square numbers. So $n^2 +n+1$ cannot be a perfect square.

There are usually two (related) strategies that are used to find solutions:

• Complete the squares and factorise using algebraic identities (usually $x^2 - y^2$). Sometimes, we can find a bound on the variables and brute force our way through. If the perfect square is factored into co-prime factors, each factor itself must be a perfect square. We will then get a system of equations to solve.
• We can use the quadratic root formula and determine when the discriminant is a perfect square.

Following Lee’s advice, I suggest you practice many problems using the techniques discussed. Let me show you some of these kicks once. If you want to see other kicks, comment below (you need a Github account to comment). I will include them.

Factoring and bounding kicks

The following British Maths Olympiad problem can be solved in multiple ways. Enjoy the kicks.

(BMO R1 2011) Find all integers $n$ for which $n^2 +20n+11$ is a perfect square.

Proof 1: We claim that $n=35,-55$ are the only solutions. We proceed by completing the square. We must solve for $n$ in the following equation: \[(n+10)^2-89=k^2.\] Now $(n+10)^2 - k^2 =89$ and thus $(n+10-k)(n+10+k)=89$ . Since $89$ is prime, we get a couple of cases. Case 1: ${n+10-k,n+10+k}={1,89}$. By adding and simplifying, we get $n=35$. Case 2: ${n+10-k,n+10+k}={-1,-89}$. By adding and simplifying, we get $n=-55$.

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Proof 2:(Abhhi and Rishon) Let $N = n^2 + 20n + 11$. We will find solutions by repeatedly bounding $N$ between two consecutive squares.

We note that for $n> 35$, since $0 < 2n-70$, by adding $n^2 + 18n + 81$ to both sides we see that $(n+9)^2 < n^2 +20n + 11$. On the other hand, $n^2 + 20n + 11 < n^2 + 20 n + 100$. Thus we have for $n > 35$, \[(n+9)^2 < n^2 + 20n + 11 < (n+10)^2.\] Since $N$ is bounded between two consecutive squares, it cannot be a perfect square. For $n=35$, we see that $N= 44^2$. For $13 < n < 35$, we similarly bound $N$ again as \[(n+8)^2 < n^2 + 20n + 11 < (n+9)^2.\] Indeed, the lower bound reduces to $4n > 53$ which holds for $n > 13$ and the upper bound simplifies to $n < 35$. So we have bounded $N$ between two consecutive squares again, and thus, N cannot be a perfect square. Now $n=13$ is not a solution since $N=440$ is a non-square number. Repeating the bounding trick, we can check that \[(n+7)^2 < n^2 + 20n + 11 < (n+8)^2\] holds for $6 < n < 13$ and conclude that $N$ is not a perfect square. Finally, if we plugin $n \in \{1,2,3,4,5,6\}$ we get $N \in \{32,55,80,107,136,167\}$. None of these values of $N$ are perfect squares. So we are done.

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Comments: We note that Proof 1 simplified because $89$ had very few factors. On the other hand, the strategy of Proof 2 is general. Try solving the problem using the quadratic equation method.

Quadratic residue kick

Next, I will show you another kick that uses quadratic residues. Watch closely.

(Croatian MO 2009) Find all positive integers $m, n$ such that $6^m+2^n+2$ is a perfect square.

Proof: We claim the only solution is $m=1,n=3.$ Let $N = 6^m+2^n+2=2(2^{m-1}3^m+2^{n-1}+1)$ be a perfect square. Since $N$ is even, if it is a square, then 4 should divide $N$. This means $2^{m-1}3^m+2^{n-1}+1$ is even. Since the expression is odd if $m,n>1$, we must have $m=1$ or $n=1$.

Case 1: In this case we consider $m=1$. This condition reduces to finding $n$ so that $N=2^{n} + 8$ is a perfect square. We know that the quadratic residues modulo 16 are only $0,1,4,9$. But if $n > 3$, then $N \equiv 8 \mod 16$ , so it cannot be a perfect square. So n can only be 1,2 or 3. Trying each of the cases, we get N as $10,12,16$. Thus $(m,n) = (1,3)$ is the only solution in this case.

Case 2: In this case we consider $n=1$. This condition reduces to finding $m$ so that $N=6^{m} +4$ is a perfect square. We know that the quadratic residues modulo $7$ are only $0,1,4,9$. So \[N = 6^m + 4 \equiv (-1)^m + 4 \equiv 5,3 \mod 7.\] Since neither are quadratic residues, we are done.

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Let me present an opportunity to practice all the kicks you have learned

Practice your kicks

1. (Kurschak MC 1953) Let n be a positive integer, and let $d$ be a positive divisor of $2n^2$. Prove that $n^2+d$ is not a perfect square. (Hint: bounding between consecutive squares.)
2. (RMO 2012) Find all positive integers $n$ such that $3^{2n} +3n^2 +7$ is a perfect square. (Hint: Bound it between consecutive squares. $2$ is the only solution)
3. (Polish MO 2000) Let $m, n$ be positive integers such that $m^2+n^2+m$ is divisible by $mn$. Prove that $m$ is a perfect square. (Hint: Use quadratic equation and focus on the discriminant.)
4. (CRMO 2001) Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is the square of an integer. (Hint: Complete squares and factorize. If $p \neq q$, then $\{3,11\}$ is the other solution.)
5. (Iranian MO 2011). Integers $a, b$ satisfy $a>b$. Also $ab-1, a+b$ are relatively prime and $ab+1, a-b$ are relatively prime. Prove that $(a+b)^2+(ab-1)^2$ is not a perfect square. (Hint: Decompose expression into co-prime factors.)
6. (Australian MO 2002) Find all prime numbers $p, q, r$ such that \[p^q+p^r\] is a perfect square. (Hint: factor out a power of p and note that we have product of relatively prime numbers is a perfect square except one pesky case.)
7. {Harder} (BMO 1994) Find the first number $n$ such that the average of the sum of the squares from $1$ to $𝑛$ (where $𝑛>1$) is a perfect square. (Ans: $337$)
8. (US TST 2008) Let $n$ be a positive integer. Prove that $n^7+7$ is not a perfect square. (looks hard, didnt try much. Let me know if you get this in the comments.)