Return to Answer

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{27} 3^5 5^2 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8-1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is much different for integers of the form $N^8-1$.

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the natural restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{27} 3^5 5^2 N^3 \approx 50,000,000,000$ should work.

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d+a-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{29} 3^8 5^4 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8+N- 1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is much different for integers of the form $N^8+N-1$ (in particular, since $N^8+N-1$ is an irreducible polynomial).

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the natural restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{29} 3^8 5^4 \approx 2,000,000,000,000,000$ should work.

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{27} 3^5 5^2 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8-1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is any lower for integers of the form $N^8+1$.

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the obvious restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{27} 3^5 5^2 N^3 \approx 50,000,000,000$ should work.

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{27} 3^5 5^2 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8-1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is much different for integers of the form $N^8-1$.

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the natural restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{27} 3^5 5^2 N^3 \approx 50,000,000,000$ should work.

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d+1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{20} 3^3 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8+1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is any lower for integers of the form $N^8+1$.

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the obvious restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{20} 3^3 \approx 10,000,000$ should work.

A better example: I believe conjecturally the equation $(a-N)^2 + (b-a^2)^2 + (c-b^2)^2 + (d-c^2)^2 + (x^2 - (d-1) y^2+1)^2 $ in the variables $a,b,c,d,x,y$, with $N$ fixed, should do the trick for most values of $N$.

We have $v(P) =2^{27} 3^5 5^2 N^3$.

The first few terms force $d= N^8$ and the last term gives the negative Pell's equation $x^2- (N^8-1)y^2=-1$. This is solvable for a majority of integers by work of Koymans and Pagano On Stevenson's Conjecture and there is no reason to suspect the proportion is any lower for integers of the form $N^8+1$.

The least solution of Pell's equation is of size roughly exp of the square root of the discriminant as long as the class group is not too large, which it is under the obvious restriction of the Cohen-Lenstra conjecture to a polynomial subsequence of discriminants.

So you just have to take $N$ large enough that exp of $N$ to the fourth power beats exp of $N$ to the third power times a rather large constant, so $N$ very roughly on the order of $2^{27} 3^5 5^2 N^3 \approx 50,000,000,000$ should work.