Question 1208797
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factorise x^4 + 4^2023.
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<pre>
Let' start writing

   {{{x^4 + 4^2023}}} = {{{x^4 + 4^3*4^2020}}} = {{{x^4 + 64*4^2020}}} = {{{x^4 + 8^2*(4^1010)^2}}} = {{{x^4 + (8*4^1010)^2}}}.


It has the form {{{x^4 + a^2}}},  where  a = {{{8*4^1010}}}.



Now we will factor  the binomial  {{{x^4 + a^2}}}  by some tricky way


      {{{x^4 + a^2}}} = {{{(x^4 + 2a*x^2 + a^2) - 2a*x^2}}} = {{{(x^2 + a)^2}}} - {{{(sqrt(2a)*x)^2}}} = 
    

    = factorize as the difference of squares = {{{(x^2 + a + sqrt(2a)x)*(x^2 + a - sqrt(2a)x)}}} = {{{(x^2 + sqrt(2a)x + a)*(x^2 - sqrt(2a)x + a)}}}.


Now substitute  a = {{{8*4^1010}}}  into this formula.  Notice that  {{{sqrt(2a)}}} = {{{sqrt(2*8*4^1010)}}} = {{{sqrt(16*4^1010)}}} = {{{4*4^505}}} = {{{4^506}}}.  
You will get


    {{{x^4 + 4^2023}}} = {{{(x^2 + 4^506*x + 8*4^1010)*(x^2 - 4^506*x + 8*4^1010)}}},


or, which is the same,


    {{{x^4 + 4^2023}}} = {{{(x^2 + 4^506*x + 2*4^1011)*(x^2 - 4^506*x + 2*4^1011)}}}.


It is the desired factorization.
</pre>

Solved.


In whole, &nbsp;it looks like a miracle.


I would say more&nbsp;: &nbsp;&nbsp;not only it looks like a miracle, &nbsp;it &nbsp;{{{highlight(highlight(IS))}}} &nbsp;a miracle.