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For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$,

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible.

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..

Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.

So they must be both even, which is the case iff $A \equiv m \mod(2)$ (provided that when $m$ is odd, $(m - 1) / B$ is even, in other words $B$ does not divide out the power of 2 dividing $m - 1$).

Furthermore from the form of $z$, $t$, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

I'm not interested in the values of integer solutions mod 6, because it seems like an artificial condition. But I dare say one could elaborate the above to cater for that also.

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$, we now investigate how to specialize this to integer solutions.

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible (given the other constraints of the problem).

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$ [*]

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..

Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.

So they must be both even, which is the case iff $A \equiv m \mod(2)$ (provided that when $m$ is odd, $(m - 1) / B$ is even, in other words $B$ does not divide out the power of 2 dividing $m - 1$).

Furthermore from the form of $z$, $t$, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

Note that the above isn't an explicit integer solution. All I have done is reduce the problem to the pair [*], to which I have a draft solution that needs checking. But if anyone else wishes to nip in first with a solution to these then obviously feel free!

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$,

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible.

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

(work in progress - temporary save made ..)

$z, t$ are integers iff $A \mid 2 L$ and $B \mid L - 2$.

Furthermore from their form, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

I'm not interested in the values of integer solutions mod 6, because it seems like an artificial condition. But I dare say one could elaborate the above to cater for that also.

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$,

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible.

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..

Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.

So they must be both even, which is the case iff $A \equiv m \mod(2)$ (provided that when $m$ is odd, $(m - 1) / B$ is even, in other words $B$ does not divide out the power of 2 dividing $m - 1$).

Furthermore from the form of $z$, $t$, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

I'm not interested in the values of integer solutions mod 6, because it seems like an artificial condition. But I dare say one could elaborate the above to cater for that also.

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$,

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $M$ we have:

$a^2 c, b(b^2 - c^2) = L M, (L - 2) M$

If $L$ is even we can replace $L, L - 2$ by $L/2, L/2 - 1$ in what follows and the same conclusion holds.

Multiplying the equations for $z$ and $t$ by $a b$, and plugging the above pair into the result gives:

$a b z = a^2 (L - 2) + b^2 L$

$a b t = - a^2 (L - 2) + b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies that $z, t$ are integers if and only if $A \mid L$ and $B \mid L - 2$.

Furthermore from their form, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

I'm not interested in the values of integer solutions mod 6, because it seems like an artificial condition. But I dare say one could elaborate the above to cater for that also.

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$,

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible.

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

(work in progress - temporary save made ..)

$z, t$ are integers iff $A \mid 2 L$ and $B \mid L - 2$.

Furthermore from their form, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

I'm not interested in the values of integer solutions mod 6, because it seems like an artificial condition. But I dare say one could elaborate the above to cater for that also.