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Show $ 3x^2 + 2 = y^2 $ has no solution in integers.

I've seen from similar problems, the idea is to reduce the equation to a congruence $ \mod{3} $ and show that the congruence $ y^2 \equiv 2 \pmod{3} $ has no solutions.

Why is one able to reduce the problem in this manner?

Jaideep Khare
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Zed1
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    Because $3$ divides $3x^2$? – Angina Seng Feb 04 '18 at 18:37
  • Because if there were a solution over the integers, it would reduce mod 3 to a solution of $y^2\equiv 2 \pmod 3$ – Ishan Levy Feb 04 '18 at 18:39
  • Why was this question put on hold? The box reads as put on hold as unclear what you're asking . Really?? I think it's clear, simple English. The OP isn't able to understand that how did we transform an equation into modular equation. So simple. Please vote to reopen this. – Jaideep Khare Feb 05 '18 at 09:49

5 Answers5

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Start from basics,

What does the representation $a \equiv b \pmod c$ mean in the first place?

Answer : It means $(b-a)$ is divisible by $c$, or in a fancy way, it's written as $$c \mid (b-a)$$

For your question, you can clearly see that if $ ~3x^2 + 2 = y^2$ is true it would imply $~ y^2-2=3x^2$. Which, therefore implies that $y^2-2$ is a multiple of $3$.

Therefore $3 \mid y^2-2 \implies y^2 \equiv 2 \pmod{3}$

So if you could prove, somehow, that this ain't possible, it would prove that the equation has no solution in integers.

Jaideep Khare
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hint: $y = 3k, 3k+1, 3k+2$ what is $y^2$ mod $3$ for each case ?

DeepSea
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Hint $3$ divides one of $y-1,y,y+1$.

If $3|y$ then $3|y^2-3x^2=2$.

If $3 \nmid y$ then $$3|y^2-1=3x^2+1$$

N. S.
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The main power of modular arithmetics relies in disproving facts. Your problem is kind of trivial, so let us study a similar but less trivial one - there are no integer solutions of $$ a^5+b^5+c^5 = 224. $$ Proof: any fifth power is either $-1,0$ or $1\pmod{11}$ and $224\equiv 4\pmod{11}$.

Finding a prime (or a set of primes) allowing to perform such disproofs might be tricky in some cases, but experience is a good teacher. As a rule of thumb, observing the involved exponents and the involved coefficients gives many leads. In $3x^2+2=y^2$ to study what happens $\!\!\pmod{3}$ comes natural.

Jack D'Aurizio
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It is same as proving that no perfect square leaves remainder 2 when divided by 3. Let n be an integer, by EDL we know that $n = 3q, 3q+1$ or $3q+2$ $\implies n^2=3(3q^2), 3(3q^2+2q) + 1$ or $3(3q^2+4q+1)+1$. So $n^2$ leaves only 0 or 1 as remainder when divided by 3. Hence, there is no perfect square which leaves 2 remainder when divided by 3.

dssknj
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