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Suppose $a$, $b$ and $c$ are three prime numbers.

How to prove that $a^2 + b^2 \neq c^2$?

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  • 1
    $\begingroup$ Welcome to Math SE, Tell us what have you tried and where are you stuck exactly. We'll be more than happy to help you $\endgroup$ – user171358 Oct 26 '14 at 16:11
  • $\begingroup$ Although many answerers have already given you essentially complete solutions for this question, in general you would learn much better if you describe what you have tried, so that people can give the appropriate guidance for you to find the solution, because reading someone else's solution never helps you gain the experience of trying and learning what techniques work and what doesn't. $\endgroup$ – user21820 Oct 26 '14 at 16:19
  • $\begingroup$ Okay, I got to this part- I assume a,b,c are all odd and that a^2+b^2 is even so when a,b,c are odd primes it's not true. $\endgroup$ – Akankshita Dash Oct 26 '14 at 16:23
  • $\begingroup$ Where I'm stuck is- I assume a=2, and b is odd= 2k+1, so a^2+b^2 is 4k^2+4k+5. How to prove this is not a perfect square?? $\endgroup$ – Akankshita Dash Oct 26 '14 at 16:24
  • $\begingroup$ @AkankshitaDash: There are a number of ways, but one very useful technique is to show that it is between two consecutive squares, because then it cannot itself be a square since $x \mapsto x^2$ is strictly increasing. Can you do this? $\endgroup$ – user21820 Oct 26 '14 at 16:30
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Hint: If $a,b$ are odd primes, $a^2 + b^2 > 2$ and is even. Hence, the only possibilities are $$ 2^2 + 2^2 = c^2 \\ 2^2 + b^2 = c^2 $$ and they are not possible either because $c - b \ge 2 \implies c^2>2^2 + b^2$.

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  • $\begingroup$ Sweet, concise and easy to understand. +1 $\endgroup$ – Asimov Oct 26 '14 at 16:09
  • $\begingroup$ That $2^2+b^2=c^2$ is not possible can also be proved by $b^2=(c-2)(c+2),$ and that $b$ is odd. $\endgroup$ – awllower Oct 26 '14 at 16:17
  • $\begingroup$ indeed. the interesting part is the first one, obviously. $\endgroup$ – mookid Oct 26 '14 at 16:23
  • $\begingroup$ Very easy to understand! But if you have to write a rigorous proof for this one, will this do? $\endgroup$ – Akankshita Dash Oct 26 '14 at 16:28
  • $\begingroup$ well, you'll have to elaborate a little. But without the details, all is here. $\endgroup$ – mookid Oct 26 '14 at 16:29
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From $a^2+b^2=c^2$ we get $a^2=c^2-b^2=(c+b)(c-b)$, i.e. a factorization of $a^2$ into two distinct factors $c+b>c-b$. The only such factorizations for the square of a prime is $a^2\cdot 1$, i.e. we conclude $c-b=1$, hence $b=2$, $c=3$. But then $a^2=5$, qea.

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  • $\begingroup$ I'm sorry, I don't get this bit- The only such factorizations for the square of a prime is a2⋅1 $\endgroup$ – Akankshita Dash Oct 26 '14 at 16:26
  • $\begingroup$ @AkankshitaDash: There are 3 possible ways to divide up the prime factors of $a^2$ into two factors. You can try all of them to see that only the one given by Hagen von Eitzen works. $\endgroup$ – user21820 Oct 26 '14 at 16:32
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The sum of two odd numbers are even, so one of the numbers must be $2$.

If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible.

If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.

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  • $\begingroup$ I'm not too familiar with what you mean by parity. Could you possibly include a resource to read more about this topic? $\endgroup$ – Brad Oct 26 '14 at 16:14
  • $\begingroup$ @brad Parity just means "odd or evenness" to say two numbers have the same parity, is to say they are either both odd, or both even. $\endgroup$ – James Oct 26 '14 at 16:18
  • $\begingroup$ +1 for avoiding the temptation of sophisticated arguments. You could not resist factoring $c^2-a^2$ though; I think my answer gives a somewhat simpler argument why $a=2$ or $b=2$ is impossible. $\endgroup$ – Marc van Leeuwen Oct 27 '14 at 6:11
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Primes are all odd expect $2$, so if $a, b, c$ don't contain $2$, $a^2 + b^2$ is even but $c^2$ is odd, then $a^2 + b^2 = c^2$ can't be true.

Of course if $c=2$, then $a^2 + b^2 = c^2$ can't be true.

If $a = 2$, then $c>b$ then $c^2 - b^2 \geq (b+2)^2 - b^2 = 4b + 4 > a^2 $

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Hint:
All Pythagorean triples can be written as:

$$ a = k\cdot(m^2 - n^2) ,\ \, b = k\cdot(2mn) ,\ \, c = k\cdot(m^2 + n^2),$$

where $m, n$, and $k$ are positive integers with $m \gt n, m − n$ odd, and with $m$ and $n$ coprime.

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On cannot have $a^2+b^2=c^2$ if $a,b,c$ are all odd. Since $2$ is the only even prime number, one of $a,b,c$ would have to be $2$. But since $2$ is the smallest prime number clearly $c$ cannot be$~2$; furthermore for any $n\geq2$ one has the inequalities $n^2<n^2+4<(n+1)^2$ showing that $n^2+2^2$ is not a square, so $a$ or $b$ cannot be $2$ either.

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because one of them must be even number.

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