On Sum of two squares

In this discussion I will prove the following statement:

If there exists two positive integers both of which can be written as sum of two squares of integers, then there product can also be written as sum of two squares of integers

Proof :

Let two positive integers xx and yy such that x=a2+b2;y=m2+n2;x=a^2+b^2;y=m^2+n^2; where all of a,b,ma,b,m and nn are integers xy=(a2+b2)(m2+n2)\Rightarrow xy=(a^2+b^2)(m^2+n^2) xy=a2m2+b2n2+a2n2+b2m2\Rightarrow xy=a^2m^2+b^2n^2+a^2n^2+b^2m^2 xy=(am+bn)22abmn+(anbm)2+2abmn\Rightarrow xy=(am+bn)^2\cancel{-2abmn}+(an-bm)^2\cancel{+2abmn} xy=(am+bn)2+(anbm)2\Rightarrow \boxed{xy=(am+bn)^2+(an-bm)^2}

Note by Zakir Husain
4 months, 1 week ago

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I have a request for you, @Zakir Husain: Can you prove this:

p1n+p2n=p3np_1^n + p_2^n = p_3^n or p1n+p2np3np_1^n + p_2^n \neq p_3^n

And p1,p2,p3p_1, p_2, p_3 are all primes...

A Former Brilliant Member - 4 months, 1 week ago

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@Yajat Shamji- What to prove?

Zakir Husain - 4 months, 1 week ago

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Read the edited comment...

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member If n>2n>2 no solution will exist see Fermat's last theorem. If n=2n=2 I have proved for that no solution exists for primes. If n=1n=1 then, p1+p2=p3p_1+p_2=p_3 In the above equation, if all of p1,p2,p3p_1,p_2,p_3 are primes, Then p3>p2;p3>p1p_3>p_2;p_3>p_1 therefore p3p_3 is greater than the 2nd2_{nd} prime p3>3\Rightarrow p_3>3. p3\Rightarrow p_3 is odd \therefore both p1,p2p_1,p_2 can't be odd. Therefore one of them must be an even prime p1=2\Rightarrow p_1=2 (Assuming p2>p1p_2>p_1). And if so 2=p3p2p12=p_3-p_2 \Rightarrow p_1 and p2p_2 are twin primes.

And according to the twin prime conjecture (not proved yet) there are infinitely many such primes.

Zakir Husain - 4 months, 1 week ago

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@Zakir Husain Your right arrow is in red - it should look like this: \rightarrow - just de-capitalise the R.

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member I have edited the comment

Zakir Husain - 4 months, 1 week ago

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@Zakir Husain Thanks. Also, can you prove for negative integers, the same thing as well?

And if all is done, make a note on it! I would appreciate it if I was the person behind the idea for the note!

@Zakir Husain

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member If you didn't care whether nn is integer or not, then just let me know

Zakir Husain - 4 months, 1 week ago

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@Yajat Shamji- Proof for n<0n<0; Let k=n;kZ+-k=n;k \in Z^+ p1n+p2n=p3n1p1k+1p2k=1p3kp_1^n+p_2^n=p_3^n \Rightarrow \frac{1}{p_1^k}+\frac{1}{p_2^k}=\frac{1}{p_3^k} p1k+p2kp1kp2k=1p3k\frac{p_1^k+p_2^k}{p_1^kp_2^k}=\frac{1}{p_3^k} p1kp2k=p3k(p1k+p2k){p_1^kp_2^k}=p_3^k(p_1^k+p_2^k) p1k=p3k(p1k+p2k)p2kp2kp3k(p12+p22){p_1^k}=\frac{p_3^k(p_1^k+p_2^k)}{p_2^k} \Rightarrow p_2^k | p_3^k(p_1^2+p_2^2) Either:p2kp3kp2p3\Rightarrow Either :p_2^k | p_3^k \Rightarrow p_2 | p_3 Or:p2k(p1k+p2k)p2kp1kp2p1Or : p_2^k | (p_1^k+p_2^k) \Rightarrow p_2^k | p_1^k \Rightarrow p_2 | p_1 Both of the above statements leads us to conclude that p2p_2 will always divide either of p3,p1p_3,p_1

\therefore All of p1,p2,p3p_1,p_2,p_3 can't be prime

Note : aba|b means 'aa divides bb' or you may translate it as b0(moda)b ≡ 0 (\mod a)

Zakir Husain - 4 months, 1 week ago

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Nice! Can you make a note on this and the positive proof...

A Former Brilliant Member - 4 months, 1 week ago

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