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Let x = 1 + ( 5 0 × 5 1 × 5 2 × 5 3 ) . Then, x 2 = 1 + ( 5 0 × 5 1 × 5 2 × 5 3 ) . Notice here that we can subtract the 1 to evoke a difference of squares on the left hand side, which would prove useful. We do this to obtain
x 2 − 1 = ( 5 0 × 5 1 × 5 2 × 5 3 ) .
Factoring the left side, we are left with ( x + 1 ) ( x − 1 ) = ( 5 0 × 5 1 × 5 2 × 5 3 )
This tells us that the product on the right hand side is a product of two numbers that differ by 2 . The closest we can pair the numbers up on the right is to do so with the pairs 5 3 × 5 0 and 5 2 × 5 1 . Note that these equal 2 6 5 0 and 2 6 5 2 respectively, so we conclude that x = 2 6 5 1 .