# Prove that for all prime numbers $\ a,b,c \ , a^{2} + b^{2} \neq c^{2}$

Question:

Prove that for all prime numbers $\ a,b,c \ , a^{2} + b^{2} \neq c^{2}$

My attempt:

Assume $\ \exists a,b,c$ prime numbers such that $\ a^{2} + b^{2} = c^{2}$.

Then $\ a^2 = c^2 - b^2 \implies a^2 = (c-b)(c+b)$.

I am not sure how to show a contradiction here. Could someone please tell me the easiest way to show a contradiction from here?

• Hint: if $a$ is prime, what are the divisors of $a^2$? After that, consider cases. One of the cases is silly -- which one? For the other case, you can solve for $b,c$ in terms of $a$. – quasi Aug 14 '17 at 2:55
• Another approach: one of the three primes must be $2$ (why?) - assume it's $a$ (clearly it can't be $c$, and $a$ and $b$ are interchangeable). Then $c^2=b^2+4$. How close can two squares be? – Steven Stadnicki Aug 14 '17 at 2:59
• Steven Stadnicki's approach is better than my approach -- quicker. – quasi Aug 14 '17 at 3:00
• Note that this is also trivial using Euclid's formula for Pythagorean triples. Euclid's formula generates all primitive triples. – Michael Lee Aug 14 '17 at 3:32
• Quasi's approach does address the question in the OP - "Could someone please tell me the easiest way to show a contradiction from here". Not sure it's possible to know what the "easiest way" is, but knowing that $a^2$ can be factored in a way other than $a\cdot a$, does lead to the contradiction. – Χpẘ Aug 14 '17 at 3:32

To continue along the approach you've started: note that by unique factorization, you either have $c-b=1$, $c+b=a^2$ or $c-b=c+b=a$; both of these are impossible for prime $b$ and $c$ — can you see why?.

Alternately, you can use a sort of parity argument: exactly one of $a$, $b$, and $c$ must be $2$ (can you see why?). It trivially can't be $c$, so suppose it's $a$; then we have $4+b^2=c^2$. Now, how close can two squares be to each other?

• How do we get a contradiction from $c -b = 1$ and $\ c+b = a^2$? – user444945 Aug 18 '17 at 3:46
• @JoshMitkitzel There isn't a contradiction per se, but if $c-b=1$ and $c$ and $b$ are both prime, there's a very short list of possibilities, and you can exhaustively check that list and show that their sum isn't a square. – Steven Stadnicki Aug 18 '17 at 5:44
• For $\ c-b=c+b=a$ we get $\ a = c \implies b = 0$, which is a contradiction right? – user444945 Aug 18 '17 at 5:46
• @JoshMitkitzel Correct, since $b$ was assumed to be prime. – Steven Stadnicki Aug 18 '17 at 5:53

Clearly the numbers $a,b,c$ can't all be odd. So let say $b$ is even and thus $2$ and $a=2x+1$ and $c=2y+1$. We get $$x^2+x+1 = y^2+y$$ which give's us contradiction since left side is odd and right even.

• (Since it took me a minute: this is after some algebraic reorganization and canceling a factor of 4 on both sides.) – Steven Stadnicki Aug 14 '17 at 21:53

For any integer $a$, $a\equiv 0,1,2(\mod 3)\Rightarrow a^2\equiv 0,1,4(\mod 3)\Rightarrow a^2\equiv 0,1(\mod 3)$

Now $a,b,c$ are prime numbers and $a^2+b^2=c^2$. First, important thing to note that $a,b,c$ are distinct.

If $a=3$, then $(c+b)(c-b)=9$. Check that $b,c$ can not prime numbers.

If $b=3$, then $(c+a)(c-a)=9$. Check that $a,c$ can not prime numbers.

If $c=3$, then $a^2+b^2\equiv 2(\mod 3)\Rightarrow c^2\equiv 2(\mod 3)$, which is impossible.

If none of $a,b,c$ is $3$, $a^2+b^2\equiv 2(\mod 3)$ and $c^2\equiv 2(\mod 3)$, which is impossible.

The conclusion is yours to decide.$\space\space\space\space\space\space\space\space\blacksquare$