Find two Pythagorean right triples A,B,C and D,E,F such that A=D and B=F, where C and F are the length of the two hypotenuses, with A ≤ B < C and, D ≤ E < F
If that is impossible, then find a pair such that A=D and abs(B-F) is minimum.
computer exploration
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 | Comment 1 of 2
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clearvars,clc
pyth=double.empty(0,3);
for t=2:100
for a=1:t-1
d=a;
b=t-a;
for f=b-1:b+1
c=sqrt(a^2+b^2);
if c==floor(c) && isreal(c)
e=sqrt(f^2-d^2);
if e==floor(e) && isreal(e) && e>0 ...
&& ~isequal([a,b,c],[d,e,f])
disp([a,b,c])
disp([d,e,f])
disp(f-b)
end
end
end
end
end
finds these pairs of triples in which B differs from F by only 1:
12 16 20
12 9 15
-1
21 28 35
21 20 29
1
40 42 58
40 9 41
-1
33 56 65
33 44 55
-1
Even when the total A + B was allowed to go to 25,000, it found no cases of zero difference between B and F.
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Posted by Charlie
on 2024-02-28 13:58:55 |
