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I've listed a few ways by which it can be proved. Please correct me if any of these are wrong. Are there other possible ways which I'm missing out?

  1. If an even number is a perfect square, it must be of the form $4k$.
  2. If an odd number is a perfect square, it must be of the form $8k+1$ .
  3. If N is a perfect square it cannot be of the form $3n+2, 5n+2, 5n+3, 7n+3, 7n+5, 7n+6$.
  4. If a number is not a square number, it can be proved so if we show that it is divisible by a prime but not by its square.
  5. (A special case of 4) If a number is not a square number, it can be proved so if we show that it is divisible by a particular prime with an odd number in the power but not an even number (greater than the odd number referred to).
Rhea
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    There are several rules to rule out that an integer is a perfect square, to prove that an integer $n$ is a perfect square, just show that $m^2=n$ holds for some integer $n$ (If there is no better way, by simply calculating the square root). – Peter May 11 '19 at 16:25
  • @Peter what if it's an expression with one or more variables? – Rhea May 11 '19 at 16:28
  • In the case of one variable it helps that in the case, the expression is a square, there must be at least one mutliple root. You can use the derivate to verify that. No idea whether the case of more variables can be handled easily in general. Do you have special examples ? – Peter May 11 '19 at 16:32
  • Necessary for a polynomial to be a perfect square is that the degree must be even. – Peter May 11 '19 at 16:36
  • https://math.stackexchange.com/questions/2118813/given-positive-integers-x-y-and-that-3x2-x-4y2-y-prove-that-x-y/2118822 The accepted answer assumes (x-y) to be an integer. I guess that simplifies. – Rhea May 11 '19 at 16:42
  • What do you mean by multiple root? – Rhea May 11 '19 at 16:44
  • This is an advanced case. It is shown that an expression is a perfect square under some condition. I thought , you had something in mind as $x^2+2xy+y^2$ (which is $(x+y)^2$) – Peter May 11 '19 at 16:44
  • If $f(x)$ has a factor $(x-a)^2$ (or a higher power) , then $a$ is a multiple root. The derivate has then this root as well. – Peter May 11 '19 at 16:45
  • Maybe this helps : https://en.wikipedia.org/wiki/Vieta_jumping – Peter May 11 '19 at 16:48
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    To make @MorganRodgers' comment more explicit, for every odd prime $p$, there are exactly $\frac{p-1}{2}$ residues $r$ which are non-squares mod $p$. Which ones these are is the content of the theory of quadratic reciprocity, and one of the main reasons we all should admire Gauß. – Torsten Schoeneberg May 12 '19 at 01:42

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