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I realize that there are similar questions to this one which already exist, I want to make sure my argumentation is sound, or if it's not, then perhaps someone can point it out to me.

Since we have that $\gcd(a,b)=1$, this implies that $1\mid a $ $\land$ $1\mid b$ $\implies$ $a=k\cdot 1$ $\land$ $b=m\cdot 1$ for $m,k\in \mathbb{Z}$. Multiplying by $a$ and $b$ respectively on both sides we get that $a^2 = ak\cdot 1$ $\land$ $b^2=bm \cdot 1$. If we let $c=ak$, $l=mb$, then we have that $a^2 = c\cdot 1$ $\land$ $b^2=l\cdot 1$ $\implies$ $1\mid a^2$ $\land$ $1\mid b^2$. This means that there exist integers $x,y$ such that $a^2x+b^2y=1$ $\implies$ $\gcd(a^2,b^2)=1$.

Am I making a leap in logic or doing something that's wrong? Please let me know.

J. W. Tanner
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prill
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2 Answers2

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It does not work and I'll explain why. First of all, avoid using $\wedge$ to mean "and", it is not easy to read and $\wedge$ is also used to mean $\gcd$ ! : $\gcd(a,b)=a\wedge b$. You did not use the fact that $\gcd(a,b)=1$ but only the fact that $1|a$ and $1|b$ (which is not a strong assumption at all). When you write that there exists $m,k\in\mathbb{Z}$ such that $a=k\cdot1$ and $b=m\cdot 1$, it is a tautology as $k=a$ and $b=m$ works. In fact you only wrote tautologies (same with $a^2=c$ and $b^2=l$ as $c=a^2$, $l=b^2$ obviously work). However, you also made a logic error : $1|a^2$ and $1|b^2$ does not mean that there exists $x,y\in\mathbb{Z}$ such that $a^2x+b^2y=1$, otherwise all integers would be coprime (take $a=b=2$ to see that it can't work). I'll give you a hint : let $d=\gcd(a^2,b^2)$, then $d|a^2$ and $d|b^2$, you can't deduce that $d|a$ and $d|b$ but it would end the proof as you would have $d|\gcd(a,b)$ and thus $d=1$, but it works if $d$ belongs in a special family of numbers.

take $d$ a prime divisor of $\gcd(a^2,b^2)$, which exists if $\gcd(a^2,b^2)\neq 1$.

Tuvasbien
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There are some flaws in your proof, for example, the statement:

$$1|a^2 \land 1|b^2 \implies a^2x+b^2y=1$$ for some integers, $x$ and $y$ is not valid.

Why don't you look at the prime factorization of $a, a^2, b, b^2$ to see what is going on?

Mohammad Riazi-Kermani
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