5
$\begingroup$

Prove that $2^{10}+5^{12}$ is composite

I need to solve this using only high school mathematics. Any ideas?

$\endgroup$
  • $\begingroup$ You mean $2^{10}+5^{12}$, right? By the way to write exponents made of more than one letter/digit use the curly braces {} after the ^, e.g. 2^{99} will yield: $2^{99}$. $\endgroup$ – Hakim Oct 17 '14 at 13:33
  • $\begingroup$ @Hakim: yes, lapsus :) $\endgroup$ – Meow Oct 17 '14 at 13:35
  • $\begingroup$ Yes, I have an idea, but it would be interesting to know what you have tried. $\endgroup$ – Mark Bennet Oct 17 '14 at 13:39
  • $\begingroup$ Only H.S. ideas? Use a calculator. it is lame, it doesn't really teach something deep, but it is what most high schooles would do. $\endgroup$ – Timbuc Oct 17 '14 at 13:39
  • 3
    $\begingroup$ @Timbuc With trial division up to fourteen thousen something? I doubt that $\endgroup$ – Hagen von Eitzen Oct 17 '14 at 13:41
15
$\begingroup$

Use the binomial formulas. From $$(2^{5}+5^{6})^2=2^{10}+2\cdot 2^5\cdot 5^6+5^{12} $$ we conclude $$ 2^{10}+5^{12}=(2^{5}+5^{6})^2-(10^3)^2= (2^{5}+5^{6}+10^3)(2^{5}+5^{6}-10^3)$$

$\endgroup$
  • $\begingroup$ That was the way I saw it too $\endgroup$ – Mark Bennet Oct 17 '14 at 13:41
  • $\begingroup$ It is even nice in decimal: $244141649 = 16657 \cdot 14657 = (15657+1000)(15657-1000)$. These are primes and so hard to find manually. $\endgroup$ – lhf Oct 17 '14 at 13:44
3
$\begingroup$

$$2^{10} + 5^{12} = (5^3)^4 + 4(2^2)^4 = 125^4 + 4\cdot4^4$$

And then the result follows from Sophie Germain's identity, ie, $$a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)$$

Yielding, $$2^{10} + 5^{12} = (5^6 + 10^3 + 2^5)(5^6 - 10^3 + 2^5) = 16657\cdot 14657$$

$\endgroup$
  • $\begingroup$ It may be that identity is not usual high school's idea. $\endgroup$ – Timbuc Oct 17 '14 at 13:39
  • $\begingroup$ Its derivation is simple polynomial factorization by grouping. You don't need to know what it's called in order to use it. $\endgroup$ – Ben Frankel Oct 17 '14 at 13:40
  • $\begingroup$ @Timbuc, Either way, I am in high school and this is the way that I would do it. $\endgroup$ – Ben Frankel Oct 17 '14 at 13:55
  • $\begingroup$ Good for you, @Ben $\endgroup$ – Timbuc Oct 17 '14 at 13:57
2
$\begingroup$

Hint:

Use the Sophie Germain identity:

$$x^4+4y^4=(x^2+2xy+2y^2)(x^2-2xy+2y^2)=((x+y)^2+y^2)((x-y)^2+y^2)$$

$$5^{12}+2^{10}=(5^3)^{4}+4\cdot (2^2)^4$$

$\endgroup$
0
$\begingroup$

$2^{10}+5^{12}=(2^5)^2+(5^6)^2$ $=(2^5+5^6)^2 - 2\cdot2^5\cdot5^6$ $=(2^5+5^6)^2 - (2\cdot5)^6$ $=(2^5+5^6)^2 - (10^3)^2$ $=(2^5+5^6 - 10^3)(2^5+5^6 + 10^3)$ $=(15657 - 1000)(15657 + 1000)$ $=14657\times 16657$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.