# Yet another integral problem

Prove that there are infinitely many perfect squares of the form

$1+2^{x^{2}}+2^{y^{2}}$

where $$(x,y) \in \mathbb{Z}$$

Note by Ankit Kumar Jain
1 year, 2 months ago

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\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

- 1 year, 2 months ago

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

- 1 year, 2 months ago