On Squarimes (Square Primes)

In this note I have mentioned that it is not possible that $p_1^2+p_2^2=p_3^2$ where $p_1,p_2,p_3$ are all primes

I have spend some more time on it and got another proof for it which I consider the best one:

For $p_1,p_2,p_3$ all of which are primes it is not possible to have $p_1^2+p_2^2=p_3^2$

Proof:

Let us assume there exists primes $p_1,p_2,p_3 | p_1^2+p_2^2=p_3^2$ $p_1^2=p_3^2-p_2^2=(p_3-p_2)(p_3+p_2)$ Now pairwise factors of $p_1^2$ are $(p_1^2,1),(p_1,p_1)$ now if $p_3-p_2=p_1\Rightarrow p_3-p_2=p_1=p_3+p_2\Rightarrow p_2=0$ which is not possible $\therefore p_3-p_2=1;p_3+p_2=p_1^2 \because p_3-p_2<p_3+p_2$ But this also leads to contradiction because according to it $p_3=p_2+1$ which is not possible for $p_3>3$

Therefore no solution exists for $p_3>3$ if $p_3=3\Rightarrow p_2=2$ but $p_3^2-p_2^2=3^2-2^2=5 \Rightarrow p_1^2=5 \Rightarrow p_1=\sqrt{5},-\sqrt{5}$ which also leads to contradiction

Therefore no squarime exists

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Thank you so much! @Zakir Husain

A Former Brilliant Member - 1 year ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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