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For finding all positive solutions of the Diophantine equation
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(1) |
we first can determine such triples $x,\,y,\,z$ which are coprime. When these are then multiplied by all positive integers, one obtains all positive solutions.
Let $(x,\,y,\,z)$ be a solution of the mentioned kind. Then the numbers are pairwise coprime, since by (1), a common divisor of two of them is also a common divisor of the third. Especially, $x$ and $y$ cannot both be even. Neither can they both be odd, since because the square of any odd number is $\equiv 1 \pmod{4}$ , the equation (1) would imply an impossible congruence $2 \equiv z^2 \pmod{4}$ . Accordingly, one of the numbers, e.g. $x$ , is even and the other, $y$ , odd.
Write (1) to the form
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(2) |
Now, both factors on the right hand side are even, whence one may denote
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(3) |
giving
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(4) |
and thus (2) reads
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(5) |
Because $z$ and $y$ are coprime and $z > y > 0$ , one can infer from (4) and (3) that also $u$ and $v$ must be coprime and $u > v > 0$ . Therefore, it follows from (5) that $$u = m^2, \quad v = n^2$$ where $m$ and $n$ are coprime and $m > n > 0$ . Thus, (5) and (4) yield
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(6) |
Here, one of $m$ and $n$ is odd and the other even, since $y$ is odd.
By substituting the expressions (6) to the equation (1), one sees that it is satisfied by arbitrary values of $m$ and $n$ . If $m$ and $n$ have all the properties stated above, then $x,\,y,\,z$ are positive integers and, as one may deduce from two first of the equations (6), the numbers $x$ and $y$ and thus all three numbers are coprime.
Thus one has proved the
Theorem. All coprime positive solutions $x,\,y,\,z$ , and only them, are gotten when one substitutes for $m$ and $n$ to the formulae (6) all possible value pairs, from which always one is odd and the other even and $m > n$ .
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- K. V¨AISÄLÄ: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).
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