Thus the above equation is always a perfect square when,

\(y^2=2\cdot (x^2-1)\)

Rewriting we get,

\(\begin{align}x^2-\dfrac{y^2}{2}=1\end{align}\)

This is similar to the pell equation
which is of the form \(x^2-ny^2=1\), where \(n\) is an integer

if we make the substitution \(\dfrac{y^2}{2}=2 z^2\), we can reduce it to the form of a pell equation in \(x,z\)

now we have \(x^2-2z^2=1 \)

we can find the fundamental solution of the pell equation by writing the sequence of convergents (\(\dfrac{h_{i}}{k_{i}}\)) to the continued
fractions of \(\sqrt{n}\) .If \((x_{1},z_{1})\) is the fundamental solution of the pell equation then \(x_{1}=h_{i}\) and \(z_{1}=k_{i}\) for some \(i\) such that \(x\) is minimal

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TopNewest\(\begin{align}1+2^{x^2}+2^{y^2}&=1+2\cdot 2^{x^2-1}+2^{y^2}\\\\ \text{now let, }y^2&=2\cdot (x^2-1)\\\\ \implies 1+2^{x^2}+2^{y^2}&=1+2\cdot 2^{x^2-1}+2^{2\cdot (x^2-1)}\\\\ &=1+2\cdot 2^{x^2-1}+\left(2^{(x^2-1)}\right)^2\\\\ &=\left(1+2^{(x^2-1)}\right)^2\end{align}\)

Thus the above equation is always a perfect square when,

\(y^2=2\cdot (x^2-1)\)

Rewriting we get,

\(\begin{align}x^2-\dfrac{y^2}{2}=1\end{align}\)

This is similar to the pell equation which is of the form \(x^2-ny^2=1\), where \(n\) is an integer

if we make the substitution \(\dfrac{y^2}{2}=2 z^2\), we can reduce it to the form of a pell equation in \(x,z\)

now we have \(x^2-2z^2=1 \)

we can find the fundamental solution of the pell equation by writing the sequence of convergents (\(\dfrac{h_{i}}{k_{i}}\)) to the continued fractions of \(\sqrt{n}\) .If \((x_{1},z_{1})\) is the fundamental solution of the pell equation then \(x_{1}=h_{i}\) and \(z_{1}=k_{i}\) for some \(i\) such that \(x\) is minimal

\(\sqrt 2=1+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+_\ddots}}}}}\)

we have \(\dfrac{h_1}{k_1}=\dfrac{3}{2}\)

\((3^2-2\cdot2^2)=1\)

we find that \((3,2)\) is the fundamental solution to the given pell equation

Once the fundamental solution is found the general solution \((x_{k},z_{k})\) can be calculated by the relation ,

\((x_{k}+z_{k}\sqrt n)=(x_{1}+z_{1}\sqrt n)^k\)

in the above case we get,

\((x_{k}+z_{k}\sqrt 2)=(3+2\sqrt 2)^k\)

similarly we have,

\((x_{k}-z_{k}\sqrt 2)=(3-2\sqrt 2)^k\),as \((x_{k}+z_{k}\sqrt 2)\times(x_{k}-z_{k}\sqrt 2)=1\)

thus we get,

\( x_{k}=\dfrac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k}{2}\\\\ z_{k}=\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{2\sqrt 2}\)

since we assumed ,\(\dfrac{y^2}{2}=2 z^2\)

we have \(y=2z\)

\(\implies y_{k}=2 z_{k}=\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{\sqrt 2}\)

Thus for all numbers of the form

\(( x_{k},y_{k})=\left(\dfrac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k}{2},\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{\sqrt 2}\right)\)

\(1+2^{x^2}+2^{y^2}\) will be a perfect square

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That is complete brilliance..Mind blowing solution...Just phenomenal..

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Post your solution guys

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