Does this count?

[Bert Bates]

Not sure if this qualifies or not...
Given:
a * a = b + c and
b = c - 1
Prove:
a * a + b * b = c * c

p.s. it's a way to generate arbitrarily large right triangles with integral sides.
[ June 02, 2003: Message edited by: Bert Bates ]

[Jim Yingst]

Given
(1) a * a = b + c
and
(2) b = c - 1
Rearrange (2):
(3) 1 = c - b
Multiply (1) by (3)
(4) (a * a) * 1 = (b + c) * (c - b)
Simplify (4):
(5) a * a = c * c - b * b
Rearrange (5):
(6) c * c = a * a + b * b
--------------------
Quod erat demonstrandum

[Joel McNary]

Given:
1:  a^2 = b + c 
2:  b = c - 1
Prove: a^2 + b^2 = c^2
3:  a^2 = b + c                  | From (1)
4:  a^2 = c-1 + c                | Substituting From (2)
5:  a^2 = 2c - 1                 | Collection of terms
6:  a^2 - c^2 = -c^2 + 2c -1     | Adding -c^2 to both sides of equation
7:  a^2 - c^2 = -1(c^2 - 2c + 1) | Factoring out -1
8:  a^2 - c^2 = -1((c-1)^2)      | Algebra stuff I remember....
9:  a^2 - c^2 = -b^2             | Substituting from (2); Distribution
10: a^2 = c^2 - b^2              | Adding c^2 to each side of eq.
11: a^2 + b^2 = c^2              | Adding b^2 to each side of eq.

[ June 02, 2003: Message edited by: Joel McNary ]

[David Weitzman]

We have
1) a*a = b + c
and
2) b = c - 1
We're looking for squares, so square 2 to get
3) b*b = c*c - 2c + 1
Add 1 and 3
4) a*a + b*b = c*c - 2c + 1 + b + c
Substitue 2 (b = c - 1) into the right side of 4
5) a*a + b*b = c*c - 2c + 1 + c - 1 + c
Simplify
6) a*a + b*b = c*c

[Rob Ross]

For what integral values of a,b, and c is this equation true?

[Jessica Sant]

a=3, b=4, c=5
a^2 + b^2 = c^2
9 + 16 = 25

[Rob Ross]

but are those the only values that work?

[Eric Pascarello]

didn'y you use A^2 + B^2 = C^2 when you were in grammer school?? It is basic gem. It works for bunch of numbers.... Plug them in for A and B and solve for C and you get your answer....

[Andy Bowes]

Also works for
5, 12 & 13
7, 24 & 25
9, 40 & 41
11, 60 & 61
etc..
Can't remember any other whole number solutions to Pythagora's equation of the top of my head but there are an inifinite number.
I also have a really nice proof of Pythagora's theorem if anyone is interested.
[ June 16, 2003: Message edited by: Andy Bowes ]

[Bert Bates]

Rob -
Those aren't the only integer sided right triangles but I don't know of a formula to generate others.

[Andy Bowes]

There is a very simple formula to get a list of right angle triangles with integer sides.
Ok start with x where x is the smallest side and an odd number.
z = ((x * x ) + 1 )/2
y = z - 1