This problem is from Bangladesh Mathematical Olympiad $2023$, The problem statement is as follows-
Prove that there is sequence of $2023$ distinct positive integers such that the sum of the squares of two consecutive integer is a perfect square itself
I can come up with such a sequence but I can't prove it. Such as-
$(2^k-1)2^0 , (2^{k-1}-1)2^k , (2^{k-2}-1)2^{k+(k+1)} , (2^{k-3}-1)2^{k+(k+1)+(k+2)}\cdots\cdots\cdots$
Where $k$ represents the number of elements in the sequence I want to create. Say, I want such a sequence with 5 numbers. Plugging in $k=5$,
$(2^5-1)2^0=31$
$(2^{5-1}-1)2^5=480$
$(2^{5-2}-1)2^9=3584$
$(2^{5-3}-1)2^{12}=12288$
$(2^{5-4}-1)2^{14}=16384$
We get the sequence $31,480,3584,12288,16384$ which obviously meet up the conditions.
I also tried to find out the $n^{th}$ term of the sequence like this:
$n^{th}$ term = ${2^{k-(n-1)}-1}[2^{(n-1)(k+1-\frac{n}{2})}]$ (based on the observed property of the sequence , some simplifications will lead to this)
Then what is left to prove is :
$\sqrt{({2^{k-(n-1)}-1}[2^{(n-1)(k+1-\frac{n}{2})}])^2+({2^{k-(n)}-1}[2^{(n)(k+1-\frac{n+1}{2})}])^2}$ AND
$\sqrt{({2^{k-(n-1)}-1}[2^{(n-1)(k+1-\frac{n}{2})}])^2+({2^{k-(n-2)}-1}[2^{(n-2)(k+1-\frac{n-2}{2})}])^2}$
Are integers, that's where I am stuck, I can't conclude because huge calculations arise with seemingly no positive deduction.
Thanks, at least for reading this:)
A different approach
All we need is the Egyptian triangle. Or the most common Pythagorean triple. Let us substitute $2023$ with $5$ for the sake of simplicity. Then the following sequence works:
$$3^5\cdot 4,\;3^4\cdot 4^2,\; 3^3\cdot 4^3,\; 3^2\cdot 4^4,\; 3\cdot 4^5$$
Now sum of squares of two consecutive numbers (say, second and third one) is $$(3^3\cdot4^2)^2\cdot(3^2+4^2)= (3^3\cdot4^2)^2\cdot5^2= (3^3\cdot4^2\cdot5)^2.$$