# Pythagorean triple

1. Show that neither $$1$$ not $$2$$ can appear in any Pythagorean triple, but that every integer $$k\geq3$$ can appear.

2. Prove that for each integer $$k$$ there are only finitely many Pythagorean triple containing $$k$$.

Help me to prove this... Thank you.

• Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you. – Kyky Nov 17 '18 at 1:55
• I have no idea...... – Nithish Kumar R Nov 17 '18 at 1:59
• If $1$ appeared in a Pythagorean triple, we would have $c^2-b^2=1$. It means that there should be two consecutive integers that are also perfect squares. Since there are no consecutive squares, $1$ cannot appear in a Pythagorean triple. You can use the same argument with $2$ and, with some modifications, with $3$ and other numbers. – rafa11111 Nov 17 '18 at 2:06

$$1^2$$ and $$2^2$$ are not sums of two positive integers squares so for any Pythagorean triple $$(x, y, z)$$ we must have $$z \ge 3$$.

Now $$x^2 < x^2 + 1 < (x + 1)^2$$ for any $$x \in N$$, so $$x^2 + 1^2$$ is not a perfect square which gives $$1 \in \{ x, y, z\}$$. i.e. $$x \ne 1, y \ne 1$$, and so $$x \ge 2, y \ge 2$$. If $$y = 2$$, then $$x^2 < x^2 +y^2 = x^2 + 4 < (x+ 1)^2$$ , since $$x \ge 2$$, and thus $$x^2 +y^2$$ is not a perfect square, so $$y \ne 2$$. Similarly $$x \ne 2$$. Thus $$\{x, y, z\} ⊆ \{3, 4, 5, 6, · · · \}$$.

Let $$k ∈ N, k \ge 3$$ any. If $$k$$ is odd, then $$k^2 = 2l + 1$$ with $$l \ge 4$$ and $$l^2 + k^2 = (l + 1)^2$$;

if $$k$$ is even, then $$k^2 = 4l$$ with $$l \ge 4$$ and $$(l − 1)^2 + k^2 = (l + 1)^2$$

Thus for any integer $$k \ge 3$$, we have a Pythagorean triple $$(x, y, z)$$ such that $$k ∈ \{x, y, z\}$$

• I think you mean "are not the difference of two positive integer squares" in the first phrase. – rafa11111 Nov 17 '18 at 2:18

Since the formula $$a^2+b^2=c^2$$ can be changed into $$a^2=c^2-b^2$$, $$a^2$$ needs to mean the difference between $$2$$ squares. You will find that $$(a+1)^2-a^2$$ is equal to $$a^2-2a+1-a^2=2a+1$$ and $$(a+2)^2-a^2=a^2+4a+4-a^2=4a+4$$. The former must be odd and the latter must be even. These are obtained by unpacking the bracket and simplifying. This is the maximum values of $$c^2$$ and $$b^2$$ of which $$a^2=c^2-b^2$$ stand true as if there were any larger then $$c^2$$ and $$b^2$$ are fractional. For $$1$$, the largest values of $$c^2$$ and $$b^2$$ is $$0^2$$ and $$1^2$$ respectively, but $$0^2$$ can't be part of a side. Again, $$2^2$$ is the difference between $$0^2$$ and $$2^2$$, but $$0^2$$ can't be used.

On to the second question. We have already proven that the above formulae show that there is a limit as to the values of $$b^2$$ and $$c^2$$, and if there are infinite possible values of $$b^2$$ and $$c^2$$ then some of the must be non-integer. So by proof of contradiction for any real integer $$k$$, there can only be a finite amount of Pythagorean triples.

We have $$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2.$$

(1). The sequence $$(2^2-1^2, 3^2-2^2, 4^2-3^2,...)$$ of differences of successive squares is the sequence $$(3,5,7,...)$$ of odd numbers $$>1$$. So if $$m$$ is odd and $$m\geq 3$$ then we can find $$b$$ such that $$(b+1)^2-b^2=m.$$ That is, $$2b+1=m.$$ So let $$b=(m-1)/2$$ and $$a=b+1.$$ Then we have $$(a^2+b^2)^2=(a^2-b^2)+(2ab)^2=m^2+(2ab)^2.$$This does not work for $$m=1$$ as the member $$2ab$$ of the triplet would be $$0.$$

(2). If $$m$$ is even and $$m\geq 4$$, let $$a=m/2$$ and $$b=1.$$ Then we have $$(a^2+b^2)^2 =(a^2-b^2)^2+(2ab)^2=(a^2-b^2)+m^2.$$This does not work for $$m=2$$ as the member $$a^2-b^2$$ of the triplet would be $$0$$.

(3). If $$x,y,z$$ are positive integers with $$x^2+y^2=z^2$$ then

(3-i). It is easy to confirm that $$1\ne z\ne 2.$$

(3-ii). We have $$y so $$y\leq z-1$$ so $$x^2=z^2-y^2\geq z^2-(z-1)^2=2z-1\geq 5$$, so $$x>2.$$ Interchanging $$x,y$$ in this, we also get $$y^2\geq 5$$ so $$y>2$$.

BTW. The general formula for ALL Pyth. triplets is $$\{k(a^2-b^2),\,2kab,\, k(a^2+b^2)\}$$ for $$a,b,k\in \Bbb N$$ with $$a>b$$.