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What is the smallest prime number n which can be formulated with all of the three following integer equations? $$u^2 + 4^2 \cdot v^2 = n$$ $$w^2 + 9^2 \cdot x^2 = n$$ $$y^2 + 11^2 \cdot z^2 = n$$

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closed as off-topic by Xander Henderson, Saad, José Carlos Santos, clathratus, Paul Frost Jan 1 at 21:54

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    $\begingroup$ The three sums can only be equal if $b=0$ but in that case $n$ is not a prime. If you mean that $a,b$ are different for each equation then you should state that better, $\endgroup$ – dxiv Dec 26 '16 at 23:25
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    $\begingroup$ I'm guessing that these are just the forms in which $n$ can be expressed - i.e it's not the same $a,b$ in all the equations. $\endgroup$ – Zestylemonzi Dec 26 '16 at 23:28
  • $\begingroup$ $127$ looks good $\endgroup$ – Euler_Salter Dec 26 '16 at 23:29
  • $\begingroup$ @Euler_Salter $127$ can't be written in the form $a^2 + 11^2 b^2$. The answer is actually $2017$ (what else?) $\endgroup$ – user399601 Dec 26 '16 at 23:30
  • $\begingroup$ I assumed that $2017$ was going to be the answer! $\endgroup$ – Zestylemonzi Dec 26 '16 at 23:31
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Using the Theorem:

If a prime can be expressed as sum of two squares then the representation is unique. Proof

Then we have that the three expressions are the same.

Then for minimize $n$ we need find a prime that can be expressed as any of this forms: $$4^2\cdot 11^2 r^2+9^2 j^2$$ or $$4^2\cdot 9^2 l^2+11^2 f^2$$ But $36^2+11^2=1417$ isn't a prime and by coincidences of the life $44^2+9^2=2017$ is a prime number. We are done.

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  • $\begingroup$ I like this theorem. $\endgroup$ – Ricardo Largaespada Dec 27 '16 at 0:15
  • $\begingroup$ I don't understand "we need find a prime such that $4^2⋅11^2r^2+9^2j^2$ or $4^2\cdot 9^2l^2+11^2f^2$". $\endgroup$ – Xam Dec 27 '16 at 4:03
  • $\begingroup$ We need a prime that can be represented as any of this forms. Did you understand? $\endgroup$ – Ricardo Largaespada Dec 27 '16 at 4:16
  • $\begingroup$ Now I do. I suggest you to edit your answer to make it more comprehensible. $\endgroup$ – Xam Dec 27 '16 at 4:20
  • $\begingroup$ I suppose that $9^211^2p^2+4^2q^2$ could have worked. But then the first term is at least $9801$. $\endgroup$ – Oscar Lanzi Nov 13 '18 at 18:57
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$$ax_1^2+bx_2^2=ax_3^2+bx_4^2=ax_5^2+bx_6^2$$

The solution is easy to write.

$$x_1=t(bp^2+ak^2)(bs^2+ak^2)$$

$$x_2=y(bp^2+ak^2)(bs^2+ak^2)$$

$$x_3=((bp^2-ak^2)t-2bpky)(bs^2+ak^2)$$

$$x_4=((bp^2-ak^2)y+2apkt)(bs^2+ak^2)$$

$$x_5=((bs^2-ak^2)t-2bsky)(bp^2+ak^2)$$

$$x_6=((bs^2-ak^2)y+2askt)(bp^2+ak^2)$$

Here the representation of 3 options, but it is easy to see that can be written in the form of a combination with any number of options.

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