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Three consecutive odd integers are such that the square of the third integer is 9 less than the sum of the squares of the first two. One solution is negative −3​, −1​, and 1. Find three other consecutive odd integers that also satisfy the given conditions

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caylus
Hello,

Let's assume a the first odd integer
a+2 the second,
a+4 the third.

(a+4)²=a²+(a+2)²-9
==> a²+8a+16=a²+a²+4a+4-9
==> a²-4a-21=0
Δ=16+4*21=10²
==> a=(4-10)/2=-3 or a=(4+10)/2=7
sol n°1=(-3;-1;1)
sol n°2=(7;9;11)


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