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

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

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

• 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: $2^{99}$. – Hakim Oct 17 '14 at 13:33
• @Hakim: yes, lapsus :) – Meow Oct 17 '14 at 13:35
• Yes, I have an idea, but it would be interesting to know what you have tried. – Mark Bennet Oct 17 '14 at 13:39
• 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. – Timbuc Oct 17 '14 at 13:39
• @Timbuc With trial division up to fourteen thousen something? I doubt that – Hagen von Eitzen Oct 17 '14 at 13:41

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

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

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

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

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

Hint:

Use the Sophie Germain identity:

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

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

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