Here's a fairly easy one.
Suppose you have a right triangle where the length of each side is a whole number. Suppose the length of one of the nonhypotenus sides is 1994. What is the length of the hypotenuse?
(Yes, ceptimus, I still need to get started on your Queen's walk :) )
A 10year old triangle

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No, there isn't; given a number a >2, we can always construct at least one Pythagorean triangle with it:Abdul Alhazred wrote: Is there any integer greater than 2 that cannot be the side of an integral pythagorean triangle?
If a is even
b = (a/2)^2  1
c = (a/2)^2 + 1
If a is odd
b = (a^2  1)/2
c = (a^2 + 1)/2

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Re: Pythagorean triangles with one side equal to 120
This problem is a bit more tedious, since we have to consider that the sides may have a common factor k; the formulas then become:Abdul Alhazred wrote:Pythagorean triangles with one side equal to 120:
If 120 is the hypotenuse, then we have a 24 times 3, 4, 5:
72, 96, 120
Otherwise there exist pythagorean triangles where one side equals 120 and the other (nonhypotenuse) side equals:
22
27
35
50
64
90
119
126
160
182
209
225
288
350
442
715
896
1197
1798
I have left out three of them.
a = k(m^2  n^2)
b = k(2mn)
c = k(m^2 + n^2)
k=1
2mn=120 mn=60= 60*1=30*2=20*3=15*4=12*5=10*6
m=60,n=1 a=3599
m=30,n=2 a=896
m=20,n=3 a=391
m=15,n=4 a=209
m=12,n=5 a=119
m=10,n=6 a=64
(Note: the sides have no common factor iff m and n are relatively prime.)
k=2
2mn=60 mn=30 =30*1 = 15*2 = 10*3 = 6*5
m=30,n=1 a=1798
m=15,n=2 a=442
m=10,n=3 a=182
m=6,n=5 a=22
k=3
2mn=40 mn=20 = 20*1 = 10*2 = 5*4
m=20,n=1 a=1197
m=10,n=2 a=288
m=5,n=4 a=27
k=4
2mn=30 mn=15 = 15*1 = 5*3
m=15,n=1 a=896
m=5,n=3 a=64
k=5
2mn=24 mn=12 = 12*1 = 6*2 = 4*3
m=12,n=1 a=715
m=6,n=2 a=160
m=4,n=3 a=35
k=6
2mn=20 mn=10 = 10*1 = 5*2
m=10,n=1 a=594
m=5,n=2 a=126
k=10
2mn=12 mn=6 = 6*1 = 3*2
m=6,n=1 a=350
m=3,n=2 a=50
The missing sides are 3599, 391 and 594.

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