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I found the following argument in a textbook that uses a number theory approach. However I do not understand the last two sentences that seems to use some number theory properties, can someone please help explain them?

Let $$ a^2 = 2b^2, \quad (a,b)=1, \quad a,b\in \mathbb{N} $$

Rearranging, $$ b^2 = (a+b)(a-b) $$

Let $p$ be a prime factor of $b$, then $$ p \mid (a+b)\quad or\quad p\mid (a-b) $$

If $p$ divides any of the above, then $p$ divides both of them, hence $p$ must divide $a$. Then $p$ would be a common divisor of $a$ and $b$, a contradiction.

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    $\begingroup$ If $p | b$ and $p | (a-b)$, say, then $p |( b + (a-b))$ so $p | a$. Basically, you need the property that if $p$ divides two integers, then it also divides their sum and differences. $\endgroup$ – Malkoun Jan 18 '20 at 5:56
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    $\begingroup$ If p is a factor of b, since $b^2 = (a+b)(a-b)$, that means that the factor p either came from $(a+b)$ or $(a-b)$, correct? Since p is a factor of b already, and hence of either $(a+b), (a-b)$ then it must also be a factor of a $\endgroup$ – Dhanvi Sreenivasan Jan 18 '20 at 6:01
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(a+b) - (a-b) =2b ,so as their difference is divisible by p(it is obviuosly divisble by b and hence p), if any one (a-b) or (a+b) then other should also be divisible , hence their sum is divisble by p , sum is 2a.

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  • $\begingroup$ Huh??? I don't see how this solves the problem. $\endgroup$ – David G. Stork Jan 18 '20 at 6:09
  • $\begingroup$ we have assumed that a and b are co primes, so if a prime factor of b ,we have proved it is a prime factor of a also and hence contradiction $\endgroup$ – aryan bansal Jan 18 '20 at 6:12
  • $\begingroup$ There is a language/grammar problem here: "we have assumed that a and b are co primes, so if a prime factor of b ,we have proved it is a prime factor of a also and hence contradiction." Huh? "so if a prime factor of b"....is what??? or so if what? is a prime factor of b??? $\endgroup$ – David G. Stork Jan 18 '20 at 6:24
  • $\begingroup$ i meant p is a prime factor of b $\endgroup$ – aryan bansal Jan 18 '20 at 9:51
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The conclusion that $p\mid(a+b)$ or $p\mid(a-b)$ comes from the property of primes that if $p\mid(rs)$ then $p\mid r$ or $p\mid s$. This, in fact, is the defining property of prime elements in a ring.

For the next sentence: if $p$ divides $a+b$ then it divides $(a+b)-2b=a-b$; if $p$ divides $a-b$ then it divides $(a-b)+2b=a+b$. The author didn't need this step to conclude that $p$ divides $a$, however; they could have simply proceeded as in Malkoun's comment: if $p$ divides $a+b$ then $p$ divides $(a+b)-b=a$; if $p$ divides $a-b$ then $p$ divides $(a-b)+b=a$.

The final sentence of the proof just uses the assumption that $p\mid b$ and the conclusion that $p\mid a$ to infer that $p$ is a common divisor of $a$ and $b$, which violates the earlier assumption $(a,b)=1$.

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Assume $\sqrt{2} = \frac{a}{b}$ with $\{ a, b \} \in \mathbb{N}$ and reduced form. Then $$2 b^2 = a^2 .$$

The number of prime factors on the left is odd; the number of prime factors on the right is even. From the fundamental theorem of arithmetic (unique prime factorization), this cannot occur.

QED

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  • $\begingroup$ My goodness! I would love to hear from the downvoter why this (which I learned in math at MIT) is not correct. $\endgroup$ – David G. Stork Dec 4 '20 at 4:10

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