Patrick D. answered • 05/07/17
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Patrick the Math Doctor
Not at all that simple, sorry.
But yes, there is a pattern of perfect squares you need to observe and understand before generalizing.
What the statement is saying is, first to choose any positive integer. Then square it.
Take the next consecutive positive integer, and square it.
Finally multiply the two consecutive positive integers together and square that result.
Adding these squares together will give you a perfect square.
In fact, it will be the square of the product of the consecutive positive integers PLUS ONE.
The first SEVERAL positive integers are listed to illustrate this pattern that you MUST observe and UNDERSTAND
before generalizing. Observing the consecutive integers I have in BOLD may help.
K=1 --> 1^2 + 2^2 + (1*2)^2 = 1 + 4 + 4 = 9 = 3^2 = (1*2+1)^2
K=2 --> 2^2 + 3^2 + (2*3)^2 = 4 + 9 + 36 = 49 = 7^2 + ( 2*3+1)^2
K=3 --> 3^2 + 4^2 + (3*4)^2 = 9 + 16 + 144 = 169 = 13^2 = ( 3*4 + 1)^2
K=4 ---> 4^2 + 5^2 + (4*5)^2 = 16 + 25 + 400 = 441 = 21^2 = ( 4*5 + 1)^2
etc...
In general k^2 + (k+1)^2 + k^2(k+1)^2 = [ k(k+1) ^2 ]^2
Try listing the next several for k=5,6,7,8... and then pick a larger value of k to convince yourself that it works.
Say for example k=10.
10^2 + 11^2 + 10^2*11^2 = 100 + 121 + 110^2 = 100 + 121 = 12100 = 12321 = 111^2 = (10*11 + 1)^2
For some positive integer K,
K^2 + (k+1)^2 + k^2*(k+1)^2 =
k^2 + k^2 + 2k + 1 + k^2*(k^2 + 2k + 1) = <--- FOIL METHOD to square the binomial
k^2 + k^2 + 2k + 1 + k^4 + 2k^3 + k^2 = <-- distributes k^2
k^4 + 2k^3 + 3k^2 + 2k + 1 <--- Label this equation ALPHA
[k(k+1) + 1]^2 = [ k^2 + k + 1 ]^2 = k^4 + 2k^3 + 3k^2 + 2k + 1
which is the same as equation ALPHA.
This proves the equation is true for any choice of positive integer k.
Patrick D.
05/07/17