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

Question

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

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

Answer 1

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

Answer 2

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

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

Answer 3

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

Answer 4

Hint:

Use the Sophie Germain identity:

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

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