Mathematical Meanderings: Even Numbers as Sums of Three Pronic Numbers

I saw this problem mentioned in a post on math.stackexchange.com:

See if you can prove that every even number is the sum of three pronic numbers (if needed, we say 0 is also pronic)

I put the problem to Gemini and got the response that "it's impossible to prove that every even number is the sum of three pronic numbers". I wasn't convinced so I consulted Grok. Here was its response.

Every even non-negative integer can indeed be expressed as the sum of three pronic numbers, where pronic numbers are of the form $n(n+1)$ for non-negative integers $n \geq 0$ (thus including 0 as the pronic number for $n=0$).
To prove this, first recall that a pronic number $p_n = n(n+1) = 2 \cdot T_n$, where $T_n = \frac{n(n+1)}{2}$ is the $n$th triangular number (with $T_0 = 0$). Let $E$ be an arbitrary even non-negative integer, so $E = 2m$ for some non-negative integer $m$. The claim is equivalent to showing that every non-negative integer $m$ can be written as the sum of three triangular numbers: $m = T_a + T_b + T_c$ for some non-negative integers $a, b, c$. This is a known theorem (attributed to Gauss): every non-negative integer is the sum of at most three triangular numbers. The proof relies on a connection to the three squares theorem and modular arithmetic, as follows. Step 1: Relate triangular numbers to squares.
Observe that $8T_n + 1 = (2n + 1)^2$. Thus, every number of the form $8T_n + 1$ is an odd perfect square, and conversely, every odd perfect square is of this form for some $n$. Step 2: Reformulate the problem.

Suppose $m = T_a + T_b + T_c$. Multiplying by 8 and adding 3 gives:$$ \begin{align} 8m + 3 &= (8T_a + 1) + (8T_b + 1) + (8T_c + 1) \\ &= (2a + 1)^2 + (2b + 1)^2 + (2c + 1)^2 \end{align} $$so $8m + 3$ is the sum of three odd squares. The converse also holds: if $8m + 3$ is the sum of three odd squares, then $m$ is the sum of three triangular numbers.
Thus, it suffices to show that every number of the form $8m + 3$ (for $m \geq 0$) is the sum of three odd squares. Step 3: Use the three squares theorem
Legendre's three squares theorem states that a positive integer can be expressed as the sum of three squares if and only if it is not of the form $4^k(8l + 7)$ for non-negative integers $k, l$. Numbers of the form $8m + 3$ are congruent to 3 modulo 8. Check if such numbers can be of the forbidden form $4^k(8l + 7)$:
For $k = 0$: $8l + 7 \equiv 7 \pmod{8}$, not 3.
For $k = 1$: $4(8l + 7) = 32l + 28 \equiv 4 \pmod{8}$, not 3.
For $k \geq 2$: $4^k(8l + 7) \equiv 0 \pmod{8}$, not 3.
Since $8m + 3$ is never of the forbidden form, every such number is the sum of three squares.

Step 4: Ensure the squares are odd
It remains to show these three squares must be odd. Consider sums of three squares modulo 8. Perfect squares modulo 8 are 0, 1, or 4:
Odd squares: $\equiv 1 \pmod{8} $
Even squares: $ \equiv 0 or 4 \pmod{8} $
The possible sums modulo 8 are 0, 1, 2, 3, 4, 5, or 6 (never 7). Specifically for $\equiv 3 \pmod{8} $
Three odd squares: $1 + 1 + 1 = 3$
Two odd, one even: $1 + 1 + 0 = 2$ or $1 + 1 + 4 = 6$.
One odd, two even: $1 + 0 + 0 = 1, 1 + 0 + 4 = 5$, or $1 + 4 + 4 = 9 \equiv 1$.
Three even: $0 + 0 + 0 = 0, 0 + 0 + 4 = 4$, $ 0 + 4 + 4 = 8 \equiv 0$, or $4 + 4 + 4 = 12 \equiv 4$.
The only combination yielding $3 \pmod{8} $ is three odd squares. Therefore, since every $8m + 3$ is the sum of three squares and $\equiv 3 \pmod{8}$, it must be the sum of three odd squares. This completes the proof. For any even $E = 2m$, express $m = T_a + T_b + T_c$, then$$ \begin{align} E &= 2T_a + 2T_b + 2T_c \\ &= p_a + p_b + p_c \end{align} $$where each $p_i $ is pronic.

This is a nice straightforward proof. I've written about pronic numbers before in posts titled:

Mathematical Meanderings

Thursday, 11 September 2025

Even Numbers as Sums of Three Pronic Numbers

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