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 p1,p2,p3p_1,p_2,p_3 all of which are primes it is not possible to have p12+p22=p32p_1^2+p_2^2=p_3^2

Proof:

Let us assume there exists primes p1,p2,p3p12+p22=p32p_1,p_2,p_3 | p_1^2+p_2^2=p_3^2 p12=p32p22=(p3p2)(p3+p2)p_1^2=p_3^2-p_2^2=(p_3-p_2)(p_3+p_2) Now pairwise factors of p12p_1^2 are (p12,1),(p1,p1)(p_1^2,1),(p_1,p_1) now if p3p2=p1p3p2=p1=p3+p2p2=0p_3-p_2=p_1\Rightarrow p_3-p_2=p_1=p_3+p_2\Rightarrow p_2=0 which is not possible p3p2=1;p3+p2=p12p3p2<p3+p2\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 p3=p2+1p_3=p_2+1 which is not possible for p3>3p_3>3

Therefore no solution exists for p3>3p_3>3 if p3=3p2=2p_3=3\Rightarrow p_2=2 but p32p22=3222=5p12=5p1=5,5p_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

Note by Zakir Husain
2 weeks, 3 days ago

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Thank you so much! @Zakir Husain

A Former Brilliant Member - 2 weeks, 3 days ago

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