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  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.

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    $\begingroup$ Welcome to MSE. Please show any working out you have done. That way, more people will be willing to help you. $\endgroup$ – Kyky Nov 17 '18 at 1:55
  • $\begingroup$ I have no idea...... $\endgroup$ – Nithish Kumar R Nov 17 '18 at 1:59
  • $\begingroup$ 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. $\endgroup$ – rafa11111 Nov 17 '18 at 2:06
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$1^2$ 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 $k ∈ \{x, y, z\}$

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

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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 $y<z$ 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 $a>b$.

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