Saturday

February 13, 2016
Posted by **-=-=-=-=-=-=-=-=-=-=-** on Thursday, July 7, 2005 at 5:14am.

Next find the sum of all the Hypotenuses from these triples you found in step 1.

Finally add together the results that you recieved in the first two steps and you'll receive the answer.

By using the method described at this website,

http://www2.math.uic.edu/~fields/puzzle/triples.html

you should be able to convince yourself that there are (1/2)*11*12 = 66 Pythagorean triples for which no side exceeds 300. The largest side will be 11^2 + 12^2 = 265 and there are 11 integer pairs for which that is true, such as (264, 23, 265) and (24, 143, 145).

The sum of the hypotenuses will be

(1^2 + 2^2) + (2^2 + 3^2) + (3^2 + 4^2) + ...((11^2 + 12^2) =

1 + 2(4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121) + 144 = 1155

The "answer" is 1155 + 66 = 1221

Apparently TchrWill has found some longer Pythagorean combination with side lengths under 300. I agree with his comments and with the website that I referenced, which tells you now to find the allowed combinations.

I have truble mith maths (AND REALLY REALLY VERY NEED HELP!)but at lest i'm REALLY good at english ! (BUT NOT THAT MUCH AT SPELLING!)

The most general formulas for deriving all integer sided right-angled Pythagorean triangles, have been known since the days of Diophantus and the early Greeks. For a right triangle with sides X, Y, and Z, Z being the hypotenuse, the lengths of the three sides of the triangle can be derived as follows: X = k(m^2 - n^2), Y = k(2mn), and Z = k(m^2 + n^2) where k = 1 for primitive triangles (X, Y, and Z having no common factor), m and n are arbitrarily selected integers, one odd, one even, usually called generating numbers, with m greater than n. The symbol ^ means "raised to the power of" such that m^2 means m squared, etc.

You can create a table of primitive triple hypotenuses by assuming values of m and n. Such a table would begin with

m............2.....3....4....5....6....7....8....9....10....11....12....13....14....15....16....17

n

1............5....10..17...26..37..50..65..82..101..122..145..170..199..226..257..290

2..................13...............................................................................................293

3........................25.................................................................................265

4..............................41...........................................................................272

5....................................61.....................................................................281

6...........................................85..............................................................292

7...............................................113...............................................274

8.....................................................145.........................................289

9.............................................................181...........................277

10...................................................................221...................296

11...........................................................................265..290

I'll let you fill in the others between these boundries. There are 87 in all..It now must be recognized that these assume k = 1. If you now consider values of k times the hypotenuses derived above, you will have many more.

Note also that the sums of each row values form a finite difference sequence with their third differences being constant. An expression can be derived for the sums of a finite difference sequence but it would probably just as easy to add them up directly.

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