Thank you!

10, 11, 12, 13, 14 -> (100, 121, 144, 169, 196) -> 365 = 365

Nice puzzle. My first thought was that it wasn't possible, but I didn't see how to prove it. Life being a cliché, I thought of a method in the shower this morning. Express the roots as k, k+1, k+2..., then square them. I thought that a bit of algebra would demonstrate a fundamental inequality, but instead it reduced to a simple quadratic equation, k² - 8k - 20 = 0, which has 10 and -2 as solutions. If you allow negative integers, the sequence -2, -1, 0, 1, 2 -> 4, 1 , 0, 1, 4 would also work, which I hadn't considered till I solved the formula.

I used a spreadsheet to study the patterns. The sum of the first three squares was greater than the last two squares when k was a large number:

121+144+169 = 434

196+225 = 421

Difference = -13

I played around with smaller numbers and the first three numbers were less than the last two numbers when k is quite small:

1+4+9 = 14

16+25+ 41

Difference = 27

Since the sum of the last two numbers will always be an odd number, the sequence has to begin with an even number.

I tried to find differences between the two sets of squares that were closer to zero but that line of attack didn’t help much.

I then thought of negative values of k, first with negative 1.

1+0+1 = 2

4+9 = 13

Difference = 11

I then explored negative 2 in the hopes of seeing a smaller difference and found a correct solution as a result.

Out of curiosity, I explored the differences for k values of -3,-4,-5, and -10, and found the following differences:

-13, -28, -45, ... -160

I like your formula approach as well. Always a great way to confirm that you’ve found all of the solutions.

🍻

I like your formula approach as well. Always a great way to confirm that you’ve found all of the solutions.

Cheers, Stephen. Actually, I had set out to prove it couldn't be done and it was a surprise to me to find the two solutions. That's the sort of thing that makes math fun.