Determine x

[AdkAdi]

Determine all [math]x[/math] such that [math]4x^5-7[/math] and [math]4x¹³-7[/math] are perfect squares.

 

[AdkAdi]

My attempt
[math]4x⁵-7=A²[/math][math]4x¹³-7=B²[/math]Subtracting both
[math]B²-A²=4x⁵(x⁸-1)[/math][math](B+A)(B-A)=4x⁵(x⁸-1)[/math][math]B+A=4X⁵[/math][math]B-A=x⁸-1[/math][math]therefore, 2B=X⁸+4X⁵-1[/math]

 

[AdkAdi]

My attempt
[math]4x⁵-7=A² 4x¹³-7=B²[/math]Subtracting both
[math]B²-A²=4x⁵(x⁸-1)[/math][math](B+A)(B-A)=4x⁵(x⁸-1)[/math][math]B+A=4X⁵[/math][math]B-A=x⁸-1[/math][math]therefore, 2B=X⁸+4X⁵-1[/math]

I do not knw how to proceed further. please help by giving a detailed solution.

 

[blamocur]

My attempt
[math]4x⁵-7=A² 4x¹³-7=B²[/math]Subtracting both
[math]B²-A²=4x⁵(x⁸-1)[/math][math](B+A)(B-A)=4x⁵(x⁸-1)[/math][math]B+A=4X⁵[/math][math]B-A=x⁸-1[/math][math]therefore, 2B=X⁸+4X⁵-1[/math]

Note that B+A=4x⁵ and B−A=x⁸−1 is not the only way to factor [imath]4x^5(x^8-1)[/imath].

My quick and dirty script found one solution, but I don't know how to prove that it is the only one. This problem does not look easy to me.

 

[AdkAdi]

Note that B+A=4x⁵ and B−A=x⁸−1 is not the only way to factor [imath]4x^5(x^8-1)[/imath].

My quick and dirty script found one solution, but I don't know how to prove that it is the only one. This problem does not look easy to me.

yeah it's certainly not easy..btw answer is x=2..I hope that would help you to figure out more about your approach..if u are able to crack the solution do share it.

 

[blamocur]

yeah it's certainly not easy..btw answer is x=2..I hope that would help you to figure out more about your approach..if u are able to crack the solution do share it.

Yeah, I got x=2 too. But no theories

 

[AdkAdi]

Yeah, I got x=2 too. But no theories

ohh..so were you doing hit and trial?

 

[blamocur]

Just ran a script for the first million values of x.

 

[blamocur]

One can prove that [imath]x[/imath] must be even

 

[AdkAdi]

okok