How do I show that we can't write $N=114^n-1$ as sum of $3$ squares for all natural number $n>2$?

Question

I run some computations in wolfram alpha, I see that we can't write :$$N=114^n-1$$ as sum of $3$ squares, then Hop someone who can show me how I do prove that we can't write $N=114^n-1$ as sum of $3$ squares for all natural number $n>2$?

Note : I have tried to use reccurrence demonstration but it's seems not work

Thank you for any help

Answer 1

as soon as $n \geq 3,$ we know that $$ 114^n = 2^n \; 3^n \; 19^n \equiv 0 \pmod 8. $$ as a result, $$ 114^n -1 \equiv 7 \pmod 8, $$ and cannot be the sum of three squares, see https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem

Answer 2

Note that $$114^1-1=2^2+3^2+10^2.$$

Answer 3

The someone you are looking for is Legendre. Google would have found him for you. See Legendre's three-square theorem

Answer 4

Hint: Working mod 4, show that if $114^n-1$ were the sum of 3 squares, then those squares would all be odd. Working mod 8, show that there are no solutions.