What is the phthgorean triplet of the greatest number is 26
Answers
Which Pythagorean triple has the number 26 in it?
Let’s first look for primitive triples. Easy: none.
Indeed a primitive triple has the form (2uv,u2−v2,u2+v2) , where u,v are coprime, not of the same parity and u>v . We’d need 26=2uv , but this implies one of the factors is 1 and the other is 13 .
Therefore the triple must be nonprimitive, that is, of the form (ka,kb,kc) , with (a,b,c) primitive.
We see that k can only be 2 or 13 . We can exclude the latter, because 2 is in no primitive triple.
Thus we know that we have to look for the primitive triples having 13 in them.
A possible case is c=u2+v2=13 , that is u=3 and v=2 . The primitive triple is (12,5,13) and so you get (24,10,26) .
The other case is b=u2−v2=13 . This means (u−v)(u+v)=13 , so u−v=1 and u+v=13 , corresponding to u=7 and v=6 . Sure enough, the triple is (84,13,85) and you end up with (168,26,170) .
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A Pythagorean triple which includes 26 must correspond to a primitive Pythagorean triple in such the corresponding side divides 26 — 1,2,13 or 26 .
It's a well-known result that all primitive Pythagorean triples are of the form
(a2−b2),2ab,(a2+b2)
with a,b integers and a>b>0 . I choose to exclude a=b as an uninteresting degenerate case.
Anyway: 1 and 2 don't work.
13 : 52+122=132 , but any others?
It's easy to enumerate the differences between 132 and smaller odd squares, to see which of them are even squares. 12,32 too small. 52 we know. 72 is too big for 132−122 , so try against 132−102=69 . 72 is too small, 92 too big, so try 132−82=101 ; 92 too small, 112 too large, so so try 132−62=133 ; 112 is too small, so try 132 ... Oops. I won't bother to grind through these explicitly in future.
132 is odd, so the only alternative is a2−b2=132 . We can have 842+132=852 . No other primitive solution can have an even leg as large as 842 . What can it have? We've got to be able to express 132 as a sum of an odd number of consecutive odd numbers, so it must be a multiple of the middle one. It's a choice between making the middle one 1 , 13 or 169 . 1 is hopeless, 169 we've done, 13 gives us 1+3+⋯+23+25 (but we don't want the even leg to be 0 ).
26 , now. This can only be the even leg, 2ab , so ab=13 , so a=13 , b=1 . But then a2−b2 is even, and so is a2+b2 , so it's not primitive; all sides are even.
So:
52+122=132 , and scale by 2
842+132=852 , and scale by 2
and these seem to be the only solutions.
Answer:
don't know
Step-by-step explanation:
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