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let $ a^{2}$ $ b^{2}$ $ c^{2}$ $ d^{2}$ be $ 4$ squares in an ap therefore $ c^{2}$ $ =$ $ 2$ $ \cdot b^{2}$ $ -$ $ a^{2}$ and $ d^{2} =3 \cdot b^{2} - 2 \cdot a^{2}$ thus $ b^{2} + d^{2} = 2 \cdot c^{2}$ this equation has no solution {trival by infinite desent}

This equation has infinitely many solutions, for example $ (b, d, c)=(1, 1, 1)$.

sorry my solution is badly wrong

A classic by Fermat ! Another solution here.