Question 182485
<font face="Times New Roman" size="+4">
PLEASE DO NOT TYPE YOUR QUESTION IN ALL CAPITALS.  IT IS THE ELECTRONIC EQUIVALENT OF SHOUTING, THEREFORE BOTH RUDE AND ANNOYING.

</font>
<font face="Times New Roman" size="+2">

If the first integer is <i>x</i>, then the next consecutive integer is <i>x</i> + 1.


Two times the square of the first integer: *[tex \Large 2x^2]


23 more than the square of the second integer: *[tex \Large (x + 1)^2 + 23]


These two expressions are equal so:


*[tex \LARGE \text{          }\math 2x^2 = (x + 1)^2 + 23]


*[tex \LARGE \text{          }\math 2x^2 = x^2 + 2x + 1 + 23]


*[tex \LARGE \text{          }\math x^2 - 2x - 24 = 0]


*[tex \LARGE \text{          }\math -6 \times 4 = -24] and


*[tex \LARGE \text{          }\math -6 + 4 = -2], so:


*[tex \LARGE \text{          }\math (x - 6)(x + 4) = 0]


Hence:


*[tex \LARGE \text{          }\math x - 6 = 0 \ \ \Rightarrow\ \ x = 6] or


*[tex \LARGE \text{          }\math x + 4 = 0 \ \ \Rightarrow\ \ x = -4]


Check:


*[tex \LARGE \text{          }\math 2(-4)^2 = 32]


*[tex \LARGE \text{          }\math (-4 + 1)^2 + 23 = 9 + 23 = 32] Checks.


*[tex \LARGE \text{          }\math 2(6)^2 = 72]


*[tex \LARGE \text{          }\math (6 + 1)^2 + 23 = 49 + 23 = 72] Checks.


So the required integers are either -4 and -3 or 6 and 7.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>