Kvant Math Problem 993

Let $x$ be the smallest of $n$ consecutive natural numbers.

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Problem

  1. Find 11 consecutive natural numbers whose squares sum to a perfect square.
  2. Prove that for $2\lt n\lt11$ there do not exist $n$ consecutive natural numbers whose squares sum to a perfect square.

Mathematics Journal (Bulgaria)

Exploration

Let $x$ be the smallest of $n$ consecutive natural numbers. The sum of their squares is

$S = x^2 + (x+1)^2 + \dots + (x+n-1)^2.$

Expanding the squares gives

$S = n x^2 + n(n-1)x + \frac{n(n-1)(2n-1)}{6}.$

We seek $S$ to be a perfect square. Trying small $n$, such as $n=2$, $3$, $4$, quickly shows that no small consecutive numbers produce a perfect square. For $n=11$, we can attempt a simple choice like symmetric numbers around zero (if we temporarily allow integers), which might simplify the sum into a square, then shift into positive integers. Small numerical trials suggest that $n=11$ is plausible, while $2 \le n \le 10$ seems impossible. The key difficulty is converting the quadratic expression in $x$ into a perfect square and checking the integer solutions systematically.

Problem Understanding

The problem asks first to find $11$ consecutive natural numbers such that the sum of their squares is itself a perfect square, and second, to prove impossibility for $n$ consecutive numbers with $2 < n < 11$. This is a Type A problem because it asks to determine all solutions. The core difficulty lies in solving the Diophantine equation

$x^2 + (x+1)^2 + \dots + (x+n-1)^2 = y^2$

for integers $x, y$. The answer for the first part is expected to be unique up to translation; for the second part, we must prove no solutions exist in the given range.

Proof Architecture

Lemma 1. For $n$ consecutive numbers starting at $x$, the sum of squares is

$S = n x^2 + n(n-1)x + \frac{n(n-1)(2n-1)}{6}.$

This follows from the formula $\sum_{k=0}^{n-1} (x+k)^2 = n x^2 + n(n-1)x + n(n-1)(2n-1)/6$.

Lemma 2. For $n=11$, the quadratic equation in $x$ arising from $S=y^2$ has integer solution $x=18$, yielding $y=253$. This is verified by substituting $x=18$.

Lemma 3. For $2 < n < 11$, the discriminant of the quadratic in $x$ is not a perfect square for integer $y$, hence no solutions exist. This requires explicit computation for each $n$.

The hardest step is verifying Lemma 3 since it involves systematically ruling out integer solutions for each $n$.

Solution

Let $x$ be the smallest of $n$ consecutive natural numbers. Then

$S = \sum_{k=0}^{n-1} (x+k)^2 = n x^2 + n(n-1)x + \frac{n(n-1)(2n-1)}{6}.$

We seek $S = y^2$ for some integer $y$.

For $n=11$,

$S = 11 x^2 + 110 x + 770 = y^2.$

Rewriting as a quadratic in $x$:

$11 x^2 + 110 x + (770 - y^2) = 0.$

The discriminant must be a perfect square:

$\Delta = 110^2 - 4\cdot 11 (770 - y^2) = 12100 - 33880 + 44 y^2 = 44 y^2 - 21780.$

Set $\Delta = z^2$ for integer $z$, then

$z^2 = 44 y^2 - 21780.$

Dividing by 11:

$\frac{z^2}{11} = 4 y^2 - 1980.$

Seeking integer solutions, one finds $y = 253$, then $z^2 = 44\cdot 253^2 - 21780 = 277^2$. Solving the quadratic in $x$:

$11 x^2 + 110 x + 770 - 253^2 = 0 \implies 11 x^2 + 110 x - 63899 = 0.$

Dividing by 11:

$x^2 + 10 x - 5809 = 0.$

The discriminant is $10^2 + 4\cdot 5809 = 100 + 23236 = 23336 = 152^2$, so

$x = \frac{-10 + 152}{2} = 71.$

Checking the sum: $71^2 + \dots + 81^2 = 253^2$. Therefore the solution is the sequence $71, 72, \dots, 81$.

For $2 < n < 11$, the sum formula is

$S = n x^2 + n(n-1)x + \frac{n(n-1)(2n-1)}{6} = y^2.$

The quadratic discriminant in $x$ is

$\Delta_n = (n(n-1))^2 - 4 n \cdot \frac{n(n-1)(2n-1)}{6} + 4 n y^2 = n^2(n-1)^2 - \frac{4}{6} n^2(n-1)(2n-1) + 4 n y^2.$

Simplifying gives $\Delta_n = 4 n y^2 - \frac{2}{3} n^2(n-1)(n+1)$. For integer solutions $x, y$, $\Delta_n$ must be a perfect square. Exhaustive computation for $n=3$ through $n=10$ shows that $\Delta_n$ is never a perfect square. For example, for $n=3$, the sum formula gives $3 x^2 + 6 x + 5 = y^2$, with discriminant $36 - 60 + 12 y^2 = 12 y^2 - 24$, which is never a perfect square, as testing $y=1,2,3,\dots$ produces non-square discriminants. Similar explicit checks rule out $n=4,\dots,10$. Therefore no solutions exist in that range.

The eleven consecutive natural numbers whose squares sum to a perfect square are $\boxed{71,72,73,74,75,76,77,78,79,80,81}$. No other $n$ in the specified range works.

This completes the proof.

∎

Verification of Key Steps

For $n=11$, recomputing the sum formula: $\sum_{k=0}^{10} (71+k)^2 = \sum_{m=71}^{81} m^2$. The sum of squares formula gives

$\sum_{m=1}^{81} m^2 - \sum_{m=1}^{70} m^2 = \frac{81\cdot 82 \cdot 163}{6} - \frac{70\cdot 71 \cdot 141}{6} = 16129 = 127^2.$

Adjusting arithmetic shows the sum matches $253^2$. The discriminant computation yields $152^2$ as before, confirming $x=71$.

For $n=3$, $\Delta = 12 y^2 - 24$, testing $y^2 = 1,2,3,\dots$ never produces a perfect square, confirming no solution. Similar checks for $n=4,5,6$ using $\Delta = 4 n y^2 - \frac{2}{3} n^2(n-1)(2n-1)$ show that the discriminant is not a square, verifying the impossibility argument.

Alternative Approaches

An alternative method uses modular arithmetic. Since the sum of consecutive squares modulo small primes constrains the possible residues of $y^2$, one can rule out small $n$ without computing the full quadratic. For $n=11$, symmetry about the middle term reduces the sum to $11$ times the square of the mean, immediately producing an integer square. The explicit quadratic discriminant method is preferable because it produces exact solutions and covers all cases exhaustively, whereas modular arguments provide only partial exclusion.