## (x^2 – 1) (y^2 – 1) = (z^2 – 1)^2

$(x^2 - 1) \, (y^2 - 1) \; = \; (z^2 - 1)^2$

An alternate form assuming   $x, y$,   and   $z$   are positive:

$x^2 \, y^2 \; - \; x^2 \; - \; y^2 \; + \; 1 \; = \; (z^2 - 1)^2$     or

$x^2 \, y^2 \; + \; 1 \; = \; x^2 \; + \; y^2 \; + \; (z^2 - 1)^2$

Here are the first few solutions:

$(17^2 - 1) \,(3^2 - 1) = (7^2 - 1)^2$
$(99^2 - 1) \,(17^2 - 1) = (41^2 - 1)^2$
$(577^2 - 1) \,(99^2 - 1) = (239^2 - 1)^2$
$(3363^2 - 1) \,(577^2 - 1) = (1393^2 - 1)^2$
$(19601^2 - 1) \,(3363^2 - 1) = (8119^2 - 1)^2$
$(114243^2 - 1) \,(19601^2 - 1) = (47321^2 - 1)^2$
$(665857^2 - 1) \,(114243^2 - 1) = (275807^2 - 1)^2$
$(3880899^2 - 1) \,(665857^2 - 1) = (1607521^2 - 1)^2$
$(22619537^2 - 1) \,(3880899^2 - 1) = (9369319^2 - 1)^2$
$(131836323^2 - 1) \,(22619537^2 - 1) = (54608393^2 - 1)^2$
$(768398401^2 - 1) \,(131836323^2 - 1) = (318281039^2 - 1)^2$
$(4478554083^2 - 1) \,(768398401^2 - 1) = (1855077841^2 - 1)^2$
$(26102926097^2 - 1) \,(4478554083^2 - 1) = (10812186007^2 - 1)^2$
$(152139002499^2 - 1) \,(26102926097^2 - 1) = (63018038201^2 - 1)^2$
$(886731088897^2 - 1) \,(152139002499^2 - 1) = (367296043199^2 - 1)^2$
$(5168247530883^2 - 1) \,(886731088897^2 - 1) = (2140758220993^2 - 1)^2$
$(30122754096401^2 - 1) \,(5168247530883^2 - 1) = (12477253282759^2 - 1)^2$
$(175568277047523^2 - 1) \,(30122754096401^2 - 1) = (72722761475561^2 - 1)^2$
$(1023286908188737^2 - 1) \,(175568277047523^2 - 1) = (423859315570607^2 - 1)^2$

math grad - Interest: Number theory
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### One Response to (x^2 – 1) (y^2 – 1) = (z^2 – 1)^2

1. John McMahon says:

The equation

$(x^2 - 1) \, (y^2 - 1) \; = \; (z^2 - 1)^2$

Is equivalent to:

$(Q_{(2n)^2} \; - \; 1) \,(Q_{(2n + 2)^2} \; - \; 1) \; = \; (Q_{(2n + 1)^2} \; - \; 1)^2$