Finding pythagoras triples when only c is known

Finding a and b when c is known

Here's what I have

str='a=%4i b=%4i c=%4i';
for c=137
    for a=1:200
        for b=1:200
            if c==sqrt(a^2+b^2)
                disp(sprintf(str,a,b))
            end
         end
    end
end
commandwindow

Answer 1

[John D'Errico]

Alternatively...

Talk to Euclid. An old guy, maybe even dead by now. He gave us a formula for Pythagorean triples.

https://en.wikipedia.org/wiki/Pythagorean_triple

Since c=137 is prime, if a triple exists, it is in lowest form, irreducible. So we can write it as

c = u^2 + v^2

a = u^2 - v^2

b = 2*u*v

for non-negative integer u,v, relatively prime.

By inspection,

137 - (1:11).^2
ans =
   136   133   128   121   112   101    88    73    56    37    16

(See? I actually used MATLAB to answer this question!) The only pair of integers that satisfy 137 = u^2 + v^2 is [11,4], thus

137 = 11^2 + 4^2 = 121 + 16

Therefore the desired Pythagorean triple is formed from

a = 11^2 - 16^2 = 105

b = 2*11*4 = 88

Of course, we can also swap a and b to get the pair of triples: [88,105,137], and [105,88,137], but they are not truly distinct. No other triples can exist for c=137.

88^2 + 105^2 == 137^2
ans =
  logical
   1

Addendum, for those who might be interested:

Perhaps a more interesting question is, does ANY pythagorean triple exist where one of the legs of the triangle is 137, thus a or b? Can such a triple exist? This goes beyond the original question where we had c==137. May one of the legs ever be 137?

In fact, pythagorean triples where one (or both) of a or b is prime tend to be rather rare. Rare as hen's teeth in fact. We can have a triple where a and c are both prime, but not both a and b. Triples where a is prime are not that common, but they do exist.

(That is not true where c would have been prime. Triples with c as prime are far more easy to find.) The reason is pretty simple. 137 is prime. As such, it has no integer factors other than 1 or 137. That means that any Pythagorean triple where one of the members is prime is an irreducible triple.

Again, Euclid tells us that we can write such a triple in the form

c = u^2 + v^2

a = u^2 - v^2

b = 2*u*v

where u and v are coprime, positive integers. That means b would never be 137, since 2 is not a factor of 137. But then, can we write 137 (a prime) in the form

137 = u^2 - v^2 = (u-v)*(u+v)

Because of unique factorization of the integers, that could ONLY EVER HAPPEN if we had

u = v + 1

so that u - v == 1. In that case we would have

137 = (v+1 - v) * (v+1 + v) = 1*(2*v+1)

so v would be 68, and u must be 69. The resulting Pythagorean triple is therefore:

pytriple = @(u,v) [max(u,v)^2 - min(u,v)^2, 2*u*v, u^2 + v^2];

pytriple(68,69)
ans =
         137        9384        9385

137^2 + 9384^2 == 9385^2
ans =
  logical
   1   

Therefore, we can prove that the only distinct Pythagorean triples with 137 as one of the members are [88,105,137] and [137,9384,9385].

Second addemdum: Note that in the above example, with a = 137, we found c as 9385, where c was not itself also prime. Can we find all Pythagorean triples where two members of the triple are prime? Is that a finite set?

By extension of what I said above, starting with a prime (exceeding 2) for a, find all triples that contain that number as a. We can then recognize that any odd prime is uniquely representable as a product only in the form

a = 1*(2*v + 1)

thus for prime a, we must have:

v = (a-1)/2

u = v + 1

Now just recognize which possible values of prime a also result in a prime for c, the longest side of the triangle?

So we can identify the set of all 32 Pythagorean triples with both a and c prime such that a is no larger than 1000.

a = setdiff(primes(1000),2)';
v = (a-1)/2;
u = v + 1;
k = isprime(u.^2 + v.^2);
[u(k).^2 - v(k).^2, 2*u(k).*v(k), u(k).^2 + v(k).^2]'
ans =
           3           4           5
           5          12          13
          11          60          61
          19         180         181
          29         420         421
          59        1740        1741
          61        1860        1861
          71        2520        2521
          79        3120        3121
         101        5100        5101
         131        8580        8581
         139        9660        9661
         181       16380       16381
         199       19800       19801
         271       36720       36721
         349       60900       60901
         379       71820       71821
         409       83640       83641
         449      100800      100801
         461      106260      106261
         521      135720      135721
         569      161880      161881
         571      163020      163021
         631      199080      199081
         641      205440      205441
         661      218460      218461
         739      273060      273061
         751      282000      282001
         821      337020      337021
         881      388080      388081
         929      431520      431521
         991      491040      491041

Note that a = 137 is NOT in that set, as expected, since 9385 is not a prime itself. But are there infinitely many such Pythagorean triples where two members of the triple are prime? This becomes a yet more interesting question...

https://primes.utm.edu/glossary/page.php?sort=PrmPythagTriples

We see on that page, Schinzel and Sierpinski's Hypothesis H: we then expect to see infinitely many triples with two prime entries. There they list the first few such triples as I list above, but no proof is apparently known fror that claim.

Regardless, it is often fun to just wander off, to go exploring in mathematics.

Addendum the third:

Just as I ran out of steam from the last exercise, did you notice anything strange fall out of the woodwork? Is there a good reason why the long side for all of those triangles seems to end with 1, at least if we ignore the first two triangles in the list? Whenever I see an incongruity like that, it just makes me wonder if there is something behind the idea. In fact, you can check that statement. All Pythagorean triples (beyond the first such pair) with two prime legs (I call them doubly prime) have the longest leg ( c ) ending with the digit 1.

Surely there would be some primes for c that do not end with 1. But primes that are also the long side of a doubly prime Pythagorean triple? Why does this seem an interesting coincidence?

Assume that a is an odd prime. Then as we have seen, a must be of the form a=2*v+1 for some integer v, and we must have u = v+1. The long side of the triple is

c = u^2 + v^2 = (v+1)^2 + v^2 = 2*v^2 + 2*v + 1

Now lets look at the value of the two sides of a Pythagorean triple that could possibly be prime, for all possible values of v, modulo 10. I can write any number v as one of the forms 10*w+(0:9).

syms w
v = 10*w + (0:9).';

[v, 2*v+1, expand(2*v.^2 + 2*v  + 1)]
ans =
[     10*w,  20*w + 1,    200*w^2 + 20*w + 1]
[ 10*w + 1,  20*w + 3,    200*w^2 + 60*w + 5]
[ 10*w + 2,  20*w + 5,  200*w^2 + 100*w + 13]
[ 10*w + 3,  20*w + 7,  200*w^2 + 140*w + 25]
[ 10*w + 4,  20*w + 9,  200*w^2 + 180*w + 41]
[ 10*w + 5, 20*w + 11,  200*w^2 + 220*w + 61]
[ 10*w + 6, 20*w + 13,  200*w^2 + 260*w + 85]
[ 10*w + 7, 20*w + 15, 200*w^2 + 300*w + 113]
[ 10*w + 8, 20*w + 17, 200*w^2 + 340*w + 145]
[ 10*w + 9, 20*w + 19, 200*w^2 + 380*w + 181]

The first column there is v. The second column is a, the third column is c. Now, exclude all rows of that array when either the second or third column would be divisible by 5, as then a or c could not be prime.

[     10*w,  20*w + 1,    200*w^2 + 20*w + 1]
[ 10*w + 4,  20*w + 9,  200*w^2 + 180*w + 41]
[ 10*w + 5, 20*w + 11,  200*w^2 + 220*w + 61]
[ 10*w + 9, 20*w + 19, 200*w^2 + 380*w + 181]

If we do exactly that, we would find in all cases the only possibilities where both a and c can simultaneously be prime and not divisible by 5 will result in mod(c,10) being 1. The first two triples in our list above ([3,4,5] and [5,12,13]) don't need to follow this pattern, because 5 is the only prime that is also trivially divisible by 5.

Having now squeezed every ounce of blood from this rock that I can possibly squeeze, I am now done. Sigh.

[dpb]

"Regardless, it is often fun to just wander off, to go exploring in mathematics."

Indeed, and thanks! for taking the time to post your digressions, John....

[Rik]

I agree, a nice read, have a +1 from me as well. ( even if it doesn't actually answers the question for a c in general ;) )

[John D'Errico]

Thx to both of you. :)

Answer 2

[Rik]

You are quite close. The code below contains some optimization in terms of selection of values to check, and the check is modified to ignore float rounding errors.

clc
str='a=%4i b=%4i c=%4i\n';
for c=137
    c2=c^2;
    for a=1:(c-1)
        a2=a^2;
        for b=a:(c-1)
            if abs(c2-(a2+b^2))<2*eps
                fprintf(str,a,b,c)
            end
         end
    end
end

[dpb]

C is the squared value, Rik...don't want to square it.

[John D'Errico]

DPB: NO. 137 is the square of NO integer. Rik is correct, though not as efficient as possible.

"If c = 137, how would I find the matching integers of 'a' and 'b' such that a^2 + b^2 = c^2?"

[dpb]

My bad; misread indeed...if just so happens that 137 = 4^2 + 11^2 and I thunk that was the target value OP was looking for. But, I see that isn't what he actually wrote, indeed.

Mea culpa! :(

[John D'Errico]

To me, it looked like one of those times whern I look in at 4:00 am, still a bit asleep. :)

[dpb]

<vbg>

Answer 3

[dpb]

Somewhat more rudimentary would be

UP=fix(sqrt(Csq-1)); % can't be bigger ignoring degenerate case of (sqrt(Csq^2),0)
for i=1:UP
  isq=i*i;
  for j=1:UP
    jsq=j*j;
    if isq+jsq==Csq
      disp([i j])
      return
    end
  end
end

Since are dealing with integers don't need to worry about the floating point roundoff, at least as long as a double can still hold exact value for Csq (~15 decimal digits).

[John D'Errico]

Nope. Here c was supplied, not c^2. Too early in the morning I think. ;-)

"If c = 137, how would I find the matching integers of 'a' and 'b' such that a^2 + b^2 = c^2?"

[dpb]

Yeah, as noted above, that 137 is 4^2 + 11^2 made me misinterpret the input as being C^2 instead of C altho OP did write otherwise...thanks.

At least I did write Csq in the Answer...how one gets it is "exercise for the Student" :)

[Rik]

Just out of curiosity: would you miss any values if you let the second loop start with i instead of 1?

(And to be honest I had already written the tolerance-safe version when I decided to cut the sqrt from the test. I anticipated sqrt might have introduced some float rounding errors.)

[dpb]

No, it's exhaustive to make the second loop begin at the position of the first, yes. Do, of course, have to start at j==i, not i+1 as you note to get the double cases. But this will always find the two factors in the order of smaller, larger for unequal a,b (again as long as no precision overflow).

I forget just how yours was written now, first, Rik...seemed to me at the time that rounding wouldn't enter into the test because the only cases that would matter were perfect squares...so as long as didn't overflow the precision it would be exact and if it did, then the rounding test wouldn't help, anyway. But, I won't claim I was necessarily correct! :)

Answer 4

[madhan ravi]

C2 = c^2;
[X,Y]=ndgrid(1:200);
XY=[X(:).^2,Y(:).^2];
XY(sum(XY,2)==C2,:) % first column is a and the second b which is squared, so if you apply sqrt() you get the values of a and b