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While 2025 may not be a prime year (we will need to wait another couple of years for that!) it more than makes up for it by being extraordinary in quite a few ways.
It is, first of all, a perfect square - the first since 1936, and the next not until 2116. And it is fairly unique in that the two halves 20 and 25 actually add to give the square root, 45!
But 2025 is most exceptional, numerically, in bringing together unexpected relationships between TRIANGLES, SQUARES and CUBES, and even ODD numbers

Introducing the Theorem of Nicomachus

Nicomachus of Gerasa (c 60AD - 120 AD) described a surprising and beautiful relationship between the sums of cubic integers and the sums of natural numbers.
The case when \(n = 9\) is of particular interest in the year 2025.
\[\sum_{k=1}^n{k^3} = (\sum_{k=1}^n{n})^2\]\[1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2\]
When n = 9 then
\[1^3 + 2^3 + 3^3 + ... + 9^3 = (1 + 2 + 3 + ... + 9)^2 = 2025\]
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Introducing the Triangular Numbers

The sum of the first \(n\) natural numbers (also described as the \(n\)-th Triangular Number) is well defined:
\[\sum_{k=1}^n {k}\]\[1 + 2 + 3 + ... + n\]\[= \frac{n\cdot (n+1)}{2}\]\[= \ 45 \ when \ n = 9\]
The SQUARE of the sum of the first \(n\) natural numbers is
\[(1 + 2 + 3 + ... + n)^2\]\[= 2025 \ when \ n = 9\]
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Square and Odd Numbers

Nicomachus also described a relationship between the SQUARE numbers and the ODD numbers:
\[\sum_{k=1}^n (2k-1) = n^2\]\[1 = 1\]\[1 + 3 = 4 = 2^2\]\[1 + 3 + 5 = 9 = 3^2\]\[1 + 3 + 5 + ... + (2n-1) = n^2\]
= 81 when n = 9
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A Proof of the Theorem of Nicomachus

One proof of the Theorem of Nicomachus begins with another unexpected relationship between the CUBIC numbers and sums of consecutive ODD numbers:
\[1^3 = 1\]
\[2^3 = 3 + 5\]\[= (2\cdot 1 + 1) + (2\cdot 1 + 3)\]\[= 2\cdot (2\cdot 1) + (1 + 3)\]\[= 2^2 + (2^2) = 2\cdot 2^2 = 2^3\]
\[3^3 = 7 + 9 + 11\]\[= (3\cdot 2 + 1) + (3\cdot 2 + 3) + (3\cdot 2 + 5)\]\[= 3\cdot (3\cdot 2) + (1 + 3 + 5)\]\[= 3\cdot 3\cdot 2 + (3^2) = 3\cdot 3^2 = 3^3\]
\[4^3 = 13 + 15 + 17 + 19\]\[= (4\cdot 3 + 1) + (4\cdot 3 + 3) + (4\cdot 3 + 5) + (4\cdot 3 + 7)\]\[= 4\cdot (4\cdot 3) + (1 + 3 + 5 + 7)\]\[= 4\cdot 4\cdot 3 + (4^2) = 4\cdot 4^2 = 4^3\]
⇒
\[1^3 + 2^3 + 3^3 + 4^3... =\]\[1 + (3 + 5) + (7 + 9 + 11) \]\[+ (13 + 15 + 17 + 19...)\]
(We have seen that the sum of the odd numbers equals \(n^2\)!)
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In general, for all integers \(n\)
\(n^3\) equals the sum of the \(n\) odd numbers starting from \(n\cdot (n-1)+1\)
\[n^3 = (n^2-n+1) + (n^2-n+3) + ... + (n^2-n+(2n-1))\]\[n^3 = \sum_{k=1}^n {n^2-n+(2k-1)}\]\[n^3 = n(n^2-n)+ \sum_{k=1}^n{(2k-1)}\]\[n^3 = n^3 - n^2 + n^2 = n^3\]
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An Alternative View

Finally, for one more unexpected connection, consider a multiplication table, up to 9 and think about ADDING all the entries, horizontally and vertically, as shown.
1 2 3 4 5 6 7 8 9 → \(1\cdot 45\) = 45
2 4 6 8 10 12 14 16 18 → \(2\cdot 45\) = 90
3 6 9 12 15 18 21 24 27 → \(3\cdot 45\) = 135
4 8 12 16 20 24 28 32 36 → \(4\cdot 45\) = 180
5 10 15 20 25 30 35 40 45 → \(5\cdot 45\) = 225
6 12 18 24 30 36 42 48 54 → \(6\cdot 45\) = 270
7 14 21 28 35 42 49 56 63 → \(7\cdot 45\) = 315
8 16 24 32 40 48 56 64 72 → \(8\cdot 45\) = 360
9 18 27 36 45 54 63 72 81 → \(9\cdot 45\) = 405
And the total of adding every element of this table?
Well, of course, it is \(45\cdot 45 = 2025\)!
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GXWeb Cubic Playground

Explore the Squared Triangular Numbers

Triangles, Squares, Cubes and Nicomachus

Introducing the Theorem of Nicomachus

Introducing the Triangular Numbers

Square and Odd Numbers

A Proof of the Theorem of Nicomachus

An Alternative View

Mathematical ToolKit

Construct your own Model with GXWeb

Behind the Scenes

1	2	3	4	5	6	7	8	9	→	\(1\cdot 45\)	=	45
2	4	6	8	10	12	14	16	18	→	\(2\cdot 45\)	=	90
3	6	9	12	15	18	21	24	27	→	\(3\cdot 45\)	=	135
4	8	12	16	20	24	28	32	36	→	\(4\cdot 45\)	=	180
5	10	15	20	25	30	35	40	45	→	\(5\cdot 45\)	=	225
6	12	18	24	30	36	42	48	54	→	\(6\cdot 45\)	=	270
7	14	21	28	35	42	49	56	63	→	\(7\cdot 45\)	=	315
8	16	24	32	40	48	56	64	72	→	\(8\cdot 45\)	=	360
9	18	27	36	45	54	63	72	81	→	\(9\cdot 45\)	=	405