Question 761695
The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers? 
I tried this one: (x+1) (x+2) and then, (x+x+1+2)= 2x+3....
and I haven't gone farther because I was looking if i have to take thirty one or square it first?
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1st: x
2nd: x+1
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x^2+(x+1)^2 = 2(x+1)^2-31
x^2 + x^2 + 2x + 1 = 2[x^2+2x+1] - 31
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2x^2 + 2x + 1 = 2x^2 + 4x + 2 -31
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2x+1 = 4x - 29
2x = 30
x = 15
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x+1 = 16
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Cheers,
Stan H.