The Infinitude of Pythagorean Triples

In the Understanding the Fermat’s Last Theorem post, I have mentioned about Pythagorean Triples.  In this post, we will show that there are infinitely many of them. We will use intuitive reasoning to prove the theorem.

For the 100th time (kidding), recall that the Pythagorean Theorem states that in a right triangle with side lengths a, b and c, where c is the hypotenuse, the equation c^2 = a^2 + b^2  is satisfied. For example, if we have a triangle with side lengths 2 and 3 units, then the hypotenuse is \sqrt{13}. The converse of the Pythagorean theorem is also true: If you have side lengths, a, b and c, which satisfies the equation above, we are sure that the angle opposite to the longest side is a right angle.

We are familiar with right triangles with integral sides. The triangle with sides (3, 4, 5) units, for instance, is a right triangle.  This is also the same with (5, 12, 13) and (8, 15, 17).  We will call this triples, the Pythagorean triples ,or geometrically, right triangles having integral side lengths.

The first thing that we can observe about the Pythagorean triples is that there are infinitely many of them. The triple (3, 4, 5), for example, can be multiplied by any positive integer to produce another Pythagorean triple. For example 2(3,4,5) = (6,8,10) is also a Pythagorean triple.  The proof is intuitively discussed below.

Theorem: There are infinitely many Pythagorean Triples.

Proof: We have discussed that two triangles that are similar are of the same shape, but not necessarily of the same size. We also know that similar triangles have congruent corresponding angles.  Therefore, if a triangle with integral side lengths (a, b, c) is right, multiplying the side lengths with any positive integer k changes only the size and not the shape.   Therefore, our new triangle with side lengths (ak, bk, ck) is also right.

Since there are infinitely many positive integers which we can substitute to k, we can therefore conclude that there are infinitely many Pythagorean Triples.

Now that we have learned that there are infinitely many of them, in the next post, we are going to discuss some interesting strategies in generating Pythagorean triples aside from the one mentioned above.

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9 thoughts on “The Infinitude of Pythagorean Triples

  1. find the minimum value of b in the equation given below so that Z is a positive integer,
    √{(b² + 1)/2}-178 = Z, where all variables are positive integers. If Z1,Z2,Z3 respectively are the next higher values after Z then find their values mathematically.

  2. Fermat’s Last Theorem can be easily proved by recognising that for n equal to or greater than 3 the binomial expansion of (p+q)^n-(p-q)^n can only have an integer nth root if p=q. This pq notation is also useful for recognising the Pythagorean Triplets for n=2 where pq has to have an integer square root.

  3. The easiest way to prove Fermat’s Last Theorem is where n is greater or equal to 3 to find the binomial expansion difference (p+q)^n-(p-q)^n
    and recognise that this expression will only have an exact integer nth root if p=q thus nullifying the second term. This pq notation is also useful for identifying the Pythagorean Triples where n=2, in that the Pythagorean Triples will only apply if pq has an integer square root.
    (p+q)^n-(p-q)^n

  4. Further to my comments in 2012, Fermat’s Last Theorem can be proved in just over 400 words by recognising that it is a problem of distinguishing between rational and irrational numbers. Where n is an integer greater than 1, the nth root of 2 is always irrational. But this irrationality can be corrected in the binomial expansion only if p equals q, otherwise the whole expression is irrational. This proves FLT.

  5. Further to my comment of 4 August 2013, if p/q equals r which is rational then 1/r being contained in the irrational 2^(1 -[1/n]) will reduce this irrational amount to another irrational amount, thus confirming Fermat’s Last Theorem.

  6. Let r=1, so that p=q. This deliberately converts the 3 term assumption of FLT into 2 terms. [2^n – 0]^(1/n) = 2 = 2^(1/n)[(n*1) + (n*3) …]^(1/n). Hence when r=1 and p=q,
    there are just 2 terms [(n*1)) + (n*3)….]^(1/n) = 2^(1-1/n).
    It also follows that if the 3 term assumption is restored,
    [(n*1)^(1/r) + (n*3) 1/r^3 …]^(1/n) will be less than 2^(1-1/n), with r greater than1
    and p greater than q, and will be less than 2 when multiplied by 2^(1/n). There can only be equality with 2 terms not the 3 terms assumed in FLT.

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