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Answered on stackexchangeAnswered on stackexchange where Kieren MacMillian linked to this MO question. Briefly: This Thue equation is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. Ljunggren proved in 1942 that the only integer solutions are $(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. [Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean triple with $a-b = \pm 1$ and $c$ a perfect square.] This implies the desired result that the only solutions of $$ Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 $$ are the known small ones with $r^2+s^2 = 1$ or $13$. The implication we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. Ljunggren's proof is a difficult application of Skolem's $p$-adic method. In the decades since then other techniques have been developed which make the solution of such an equation routine; for example this is how gp can calculate almost instantaneously

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

in response to the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

But none of these techniques is elementary either, and as far as I know the problem of finding an elementary proof of Ljunggren's theorem remains open.

Answered on stackexchange where Kieren MacMillian linked to this MO question. Briefly: This Thue equation is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. Ljunggren proved in 1942 that the only integer solutions are $(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. [Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean triple with $a-b = \pm 1$ and $c$ a perfect square.] This implies the desired result that the only solutions of $$ Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 $$ are the known small ones with $r^2+s^2 = 1$ or $13$. The implication we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. Ljunggren's proof is a difficult application of Skolem's $p$-adic method. In the decades since then other techniques have been developed which make the solution of such an equation routine; for example this is how gp can calculate almost instantaneously

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

in response to the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

But none of these techniques is elementary either, and as far as I know the problem of finding an elementary proof of Ljunggren's theorem remains open.

Answered on stackexchange where Kieren MacMillian linked to this MO question. Briefly: This Thue equation is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. Ljunggren proved in 1942 that the only integer solutions are $(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. [Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean triple with $a-b = \pm 1$ and $c$ a perfect square.] This implies the desired result that the only solutions of $$ Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 $$ are the known small ones with $r^2+s^2 = 1$ or $13$. The implication we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. Ljunggren's proof is a difficult application of Skolem's $p$-adic method. In the decades since then other techniques have been developed which make the solution of such an equation routine; for example this is how gp can calculate almost instantaneously

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

in response to the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

But none of these techniques is elementary either, and as far as I know the problem of finding an elementary proof of Ljunggren's theorem remains open.

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Noam D. Elkies
Noam D. Elkies
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Answered on stackexchange where Kieren MacMillian linked to this MO question. Briefly: This Thue equation is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. Ljunggren proved in 1942 that the only integer solutions are $(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. [Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean triple with $a-b = \pm 1$ and $c$ a perfect square.] This implies the desired result that the only solutions of $$ Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 $$ are the known small ones with $r^2+s^2 = 1$ or $13$. The implication we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. Ljunggren's proof is a difficult application of Skolem's $p$-adic method. In the decades since then other techniques have been developed which make the solution of such an equation routine; for example this is how gp can calculate almost instantaneously

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

in response to the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

But none of these techniques is elementary either, and as far as I know the problem of finding an elementary proof of Ljunggren's theorem remains open.

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