Kvant Math Problem 316

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Problem

Prove that the sum of the squares of $k$ consecutive natural numbers cannot be a perfect square if $k$ equals 3, 5, 7, or 9.
Find 11 consecutive natural numbers whose sum of squares is a perfect square.

E. G. Gotman

Exploration

Consider the sum of squares of $k$ consecutive natural numbers beginning at $n$, expressed as

$S_k(n) = n^2 + (n+1)^2 + \dots + (n+k-1)^2 = \sum_{i=0}^{k-1} (n+i)^2.$

Expanding this yields

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

This formula allows investigation of small $k$. For $k=3$, the sum is $3n^2 + 6n + 5$, which is never a perfect square for natural $n$ because its value modulo 3 is either 2 or 0, which cannot be a square modulo 3. Similar analysis modulo small integers, particularly modulo 3 and modulo 4, can help exclude $k=5,7,9$. For $k=11$, testing small integers or attempting to center the numbers symmetrically around a mean may produce a sum that is a perfect square. The crucial difficulty is showing nonexistence of perfect squares for small odd $k$; naive trial fails, so modular arithmetic provides the essential constraint.

Problem Understanding

The first part requires proving that for $k=3,5,7,9$, the sum of squares of $k$ consecutive natural numbers is never itself a perfect square. This is a Type B problem, a pure proof of impossibility. The second part asks for an explicit example of 11 consecutive natural numbers whose sum of squares forms a perfect square, a Type D problem, requiring construction.

The core difficulty for the first part lies in showing that for odd $k$, the sum of squares, expressed as a quadratic polynomial in $n$, never attains a perfect square. For the second part, the challenge is to find numbers whose sum of squares simplifies to a square, likely by centering numbers symmetrically and exploiting sums of symmetric sequences.

Proof Architecture

Lemma 1: For odd $k$, the sum of squares of $k$ consecutive natural numbers can be expressed as a quadratic in $n$ with integer coefficients. The polynomial form follows from the expansion of $(n+i)^2$ and summing arithmetic series.

Lemma 2: For $k=3$, modulo 3 arithmetic shows the sum is never a perfect square. The sum modulo 3 produces residues incompatible with quadratic residues modulo 3.

Lemma 3: For $k=5$, modulo 5 arithmetic excludes perfect squares. The sum modulo 5 is shown to yield residues not in the set of quadratic residues modulo 5.

Lemma 4: For $k=7$ and $k=9$, a similar argument with modulo 7 and modulo 3 arithmetic shows impossibility. Quadratic residues modulo small primes suffice to exclude solutions.

Lemma 5: To find 11 consecutive numbers whose sum of squares is a perfect square, center the sequence symmetrically around $n$, express the sum in terms of $n$, and solve the resulting quadratic for integer $n$ such that the sum is a perfect square. Verification of integer solutions completes the construction.

The hardest part is Lemma 4, where careless handling of congruences may miss certain residues or overlook periodicity.

Solution

Let $S_k(n) = \sum_{i=0}^{k-1} (n+i)^2$. Expanding gives

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

For $k=3$,

$S_3(n) = 3 n^2 + 6 n + 5.$

Consider $S_3(n)$ modulo 3. Every perfect square is congruent to 0 or 1 modulo 3. The sum reduces modulo 3:

$S_3(n) \equiv 3n^2 + 6n + 5 \equiv 0 + 0 + 2 \equiv 2 \pmod{3}.$

Since 2 is not a quadratic residue modulo 3, $S_3(n)$ is never a perfect square.

For $k=5$,

$S_5(n) = 5 n^2 + 20 n + 55.$

Reducing modulo 5 gives

$S_5(n) \equiv 5 n^2 + 20 n + 55 \equiv 0 + 0 + 0 \equiv 0 \pmod{5}.$

Thus any perfect square solution must be divisible by 5. Let $S_5(n) = 5 m^2$. Substituting, we obtain

$5 n^2 + 20 n + 55 = 5 m^2 \implies n^2 + 4 n + 11 = m^2.$

Then

$n^2 + 4 n + 11 - m^2 = 0 \implies n^2 + 4 n + (11 - m^2) = 0.$

The discriminant of this quadratic in $n$ must be a perfect square:

$\Delta = 16 - 4(11 - m^2) = 16 - 44 + 4 m^2 = 4 m^2 - 28 = 4(m^2 - 7).$

This discriminant must be a perfect square, so $4(m^2 - 7) = k^2$ for some integer $k$, implying $k$ even, $k = 2 t$, and $m^2 - 7 = t^2$, hence

$m^2 - t^2 = 7 \implies (m-t)(m+t) = 7.$

The integer factor pairs of 7 are $(1,7)$ and $(7,1)$ and their negatives. Solving $m-t=1$, $m+t=7$ gives $m=4$, $t=3$; $m-t=7$, $m+t=1$ gives $m=4$, $t=-3$; $m-t=-1$, $m+t=-7$ gives $m=-4$, $t=3$; $m-t=-7$, $m+t=-1$ gives $m=-4$, $t=-3$. Each case gives $n = (-4 \pm \sqrt{\Delta})/2 = (-4 \pm 2 t)/2 = -2 \pm t$, yielding $n = 1$ or $-5$, $n=-5$ or $1$, etc. Only positive $n$ matters, so $n=1$. Then $S_5(1) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$, not a perfect square. No solution exists.

For $k=7$,

$S_7(n) = 7 n^2 + 42 n + 91.$

Modulo 7, $S_7(n) \equiv 0 + 0 + 0 \equiv 0 \pmod{7}$. Let $S_7(n) = 7 m^2$, then $n^2 + 6 n + 13 = m^2$. Discriminant $36 - 4 (13 - m^2) = 36 - 52 + 4 m^2 = 4 m^2 - 16$. Setting $4 m^2 - 16 = k^2$, $k$ even, $k=2 t$, gives $m^2 - t^2 = 4$, leading to $(m-t)(m+t) = 4$, factor pairs $(1,4),(2,2)$, etc. Each yields $n$ negative or non-integer. No solution in natural numbers.

For $k=9$, $S_9(n) = 9 n^2 + 72 n + 204$, modulo 3 gives $0 + 0 + 0 \equiv 0$, set $S_9(n) = 9 m^2$, $n^2 + 8 n + 22 = m^2$. Discriminant $64 - 88 + 4 m^2 = 4 m^2 - 24$. $4 m^2 - 24 = k^2$, $k = 2 t$, $m^2 - t^2 = 6$, factor pairs of 6 produce non-integer $n$. No solution exists.

For 11 consecutive numbers, let them be $n-5, n-4, \dots, n+5$, so the sum of squares is

$\sum_{i=-5}^{5} (n+i)^2 = \sum_{i=-5}^{5} (n^2 + 2 n i + i^2) = 11 n^2 + 2 n \sum_{i=-5}^{5} i + \sum_{i=-5}^{5} i^2 = 11 n^2 + 0 + 110 = 11 n^2 + 110.$

We seek $11 n^2 + 110 = m^2$, i.e., $m^2 - 11 n^2 = 110$. Factor $110 = 11 \cdot 10$, try $n=5$, then $11 \cdot 25 + 110 = 385 = 19^2 + 6$, adjust. Try $n=0$, $S = 110$, not a square. Trying $n=1$, $11 +