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Is there any proof or counter-proof of Mohanty's conjecture (1988) in the litterature:

The numbers n, n + 6, and n + 12 cannot be expressed simultaneously as sum of two squares.

techera1
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  • I don't understand a kind of Pythagorean triples? $N^2+(N+6)^2=(N+12)^2$ For a similar shape is there a formula. http://math.stackexchange.com/questions/5540/k2k12-being-a-perfect-square-for-infinitely-many-k/700126#700126 – individ Feb 03 '17 at 04:17
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    It means that there exist $N$ such that the three integers $N$, $N+6$ and $N+12$ can be writen as sum of two squares. – techera1 Feb 03 '17 at 07:14
  • I still don't understand. Hard is normal and okay to ask a question? Or 100 times it is necessary to ask again? It is necessary to solve such a system? $$N=a^2+b^2$$ $$N+6=c^2+q^2$$ $$N+12=x^2+y^2$$ – individ Feb 03 '17 at 09:44
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    If I understood your question then N=18 is one solution since 18^2+24^2=30^2. – user25406 Feb 03 '17 at 18:55
  • I edited the question to be more clear. – techera1 Feb 03 '17 at 21:28
  • that is not difficult to solve. you just setup your equations for the general case with your n as a parameter. you will have 3 quadratic and if the conjecture is wrong, then at least one quadratic will not have integer roots. You can use this algorithm to do that: http://math.stackexchange.com/questions/2118459/sum-of-two-squares-of-an-integer-n-the-simplest-algorithm?noredirect=1#comment4356705_2118459 – user25406 Feb 04 '17 at 01:47
  • What was edited, again it is not clear? If you need to solve the system that I mentioned. There are infinitely many solutions. And generally what there are Pythagorean triples? – individ Feb 04 '17 at 04:52
  • I think it has to do with the fact that if an integer is of the form 4k+1, then adding 6 to it makes it of the form 4k+3, which usually do not admit a decomposition as a sum of 2 squares. But adding 12 does not change the form ( ex 41 + 6 = 47 = 411 +3 but 41 + 12 = 53 = 413 + 1). So while 41 and 53 can be written as a sum of 2 squares, 47 cannot. – user25406 Feb 04 '17 at 20:16
  • There are plenty of cases. For example: $4=2^2+0^2, 10=3^2+1^2, 16=4^2+0^2$; or $20=4^2+2^2, 26=5^2+1^2, 32=4^2+4^2$. Did Mohanty impose some other condition not mentioned in the question? – Rosie F Mar 10 '23 at 16:03

1 Answers1

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For the system of equations.

$$\left\{\begin{aligned}&N=c^2+q^2\\&N+T=a^2+b^2\\&N+2T=x^2+y^2\end{aligned}\right.$$

Lay on multipliers. $T=2ps$ Solutions written in this form.

$$c=T+k^2+k(p+s-2)-p-s+\frac{1}{2}$$

$$q=T+k^2+k(p+s)-\frac{1}{2}$$

$$a=T+k^2+k(p+s-1)-p+\frac{1}{2}$$

$$b=T+k^2+k(p+s-1)-s+\frac{1}{2}$$

$$x=T+k^2+k(p+s-1)-p-s+\frac{1}{2}$$

$$y=T+k^2+k(p+s-1)+\frac{1}{2}$$

$k - $ Any whole number. It is seen that solutions in integers there is not only for $T=6$ but for any other integer.

This formula will be better ....

Decompose the number $T$ in two different ways. $T=2ps=kt$

$$c=kn^2+(2k+s-p)n+s-p+\frac{t+k}{2}$$

$$q=kn^2+(s-p)n+\frac{t-k}{2}$$

$$a=kn^2+(k+s-p)n+s-p+\frac{t+k}{2}$$

$$b=kn^2+(k+s-p)n+\frac{t+k}{2}$$

$$x=kn^2+(k+s-p)n+s+\frac{t+k}{2}$$

$$y=kn^2+(k+s-p)n-p+\frac{t+k}{2}$$

individ
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  • How did you derive these identities ? – Sandeep Silwal Mar 28 '17 at 16:50
  • It seems like you need $t+k$ to be even, for your variables to be integers. If so, this cannot be achieved with $T=6$. This is why the "odd" case allows for a solution, where we're guaranteed that $t, k$ have the same parity. – Calvin Lin Jun 17 '17 at 00:49