MHB Pythagorean Theorem: Unexpected Finding

AI Thread Summary
The discussion explores an unexpected pattern discovered while experimenting with the Pythagorean theorem using combinations of integers up to 10. The user noted that the square roots of the sums of squares exhibit an increasing sequence of differences: 3, 5, 7, 9, and 11, which themselves increase by 2. This observation leads to a deeper inquiry into the relationships between successive equations and their differences. Participants in the discussion analyze the mathematical implications of these findings, particularly focusing on the nature of the differences between successive sums of squares. The conversation emphasizes the intriguing connections within the Pythagorean theorem and invites further exploration of these patterns.
Angel11
Messages
11
Reaction score
0
Hello, today i was playing around with the pythagorean theorem and found out something that i can't really explaing or atleast explain it with probably a false answer. So i was putting every possible combination with the max digit of 10. For example 1^2+1^2=\sqrt{2}, 1^2+2^2=\sqrt{5}... 10^2+1^2=\sqrt{101},10^2+2^2=\sqrt{104}. And then i found out that the square roots of the resolts are increasing by 3,5,7,9,11 and then i found out that those number are increasing by 2. so something like "1^2+2^2=\sqrt{(1^2+1^2)}+\sqrt{3} and then 1^2+3^2=\sqrt{(1^2+2^2)}+\sqrt{5}" so any idea how it works? my guess is WAY off i thought about it more and it is awful so i would appreciate anyones response.
 
Mathematics news on Phys.org
If I understand correctly, you have observed for example:

$$1^2+0^2=1$$

$$1^2+1^2=2$$

$$1^2+2^2=5$$

$$1^2+3^2=10$$

And you've seen that the difference between successive equations are the sequence of odd natural numbers. Let's look at the difference between two successive equations in general:

$$1^2+n^2=n^2+1$$

$$1^2+(n+1)^2=n^2+2n+2$$

What do we get when we subtract the former from the latter?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
8
Views
3K
Replies
2
Views
1K
Replies
2
Views
1K
3
Replies
105
Views
6K
Replies
4
Views
10K
Back
Top