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The main question is : Is $545^4 + 4^{545}$ a prime number? Explain your answer.

My approach : I tried writing the expression as, $$545^4 + 4*4^{544}$$ Thus we get, $$545^4 + 4*{(4^{136})}^4$$

I can't proceed any further. Is there some obvious thing or concept I'm missing? This question is a base-level olympiad question, thus there must be some short solution to this. I will appreciate if you can give a detailed answer, along with a shortcut if you have one, so that I understand the concept thoroughly. Thanks!

Akshar Gandhi
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2 Answers2

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We can go through this using primitive steps. (Proceeding after the two steps given in my approach) Let $545=x$ and $4^{136}=y$. Thus, the expression becomes : $$x^4+4y^4$$ Adding and subtracting by $4x^2y^2$, we get, $${(x^2+2y^2)}^2-{(2xy)}^2$$ This can be written as : $$(x^2+2y^2+2xy)(x^2+2y^2-2xy)$$

Now, since there are two factors to the original expression excluding $1$ and itself, the expression $545^4+4^{545}$ is not a prime.

Akshar Gandhi
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    You need to also show that the second factor, of the form $x^2 + 2y^2 - 2x^2 y^2$, is not equal to $1$ for the particular choice of $x$ and $y$. – heropup Jul 30 '16 at 05:57
  • Even if it is 1, the other factor itself cannot reduce to $x^4+4y^4$, thus it is surely going to have 3 factors and not be prime. Good enough? – Akshar Gandhi Jul 30 '16 at 06:01
  • Same as the comment from @daruma – Gerry Myerson Jul 30 '16 at 06:02
  • @daruma gave me the trick, so I saved everyone else's time by answering my own question. – Akshar Gandhi Jul 30 '16 at 06:03
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    @AksharGandhi Note that $5 = 3^2 - 2^2 = (3-2)(3+2)$, but this factorization does not prove $5$ is composite, even though the general statement $a^2 - b^2 = (a-b)(a+b)$ is a factorization. This is why you must take care to establish that both resulting factors are nontrivial. Incidentally, your factorization should read $$(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2).$$ You've mistakenly squared the $xy$ terms. – heropup Jul 30 '16 at 06:17
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    @heropup $x^2-2xy+2y^2=1 \implies x=y \pm \sqrt{1-y^2}$ that is only possible when $-1 \leq y \leq 1$. – Ng Chung Tak Jul 30 '16 at 07:59
  • @NgChungTak Exactly! That's all you need to justify it, but it still needs to be said. – heropup Jul 30 '16 at 14:02
  • Okay @heropup, I get your point, but I don't really understand what to edit in the answer – Akshar Gandhi Jul 30 '16 at 16:37
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    @AksharGandhi just argue that $x^2 + y^2 - 2xy \ge 0$, so $x^2 + 2y^2 - 2xy \ge y^2 > 1$ and you are done. –  Jul 30 '16 at 17:05
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$$ x^2 - 2 x y + 2 y^2 = (x-y)^2 + y^2 \geq y^2 $$

Will Jagy
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