TargetIIT

Rohan heres my sol its quite diff from wat ur expecting

case 1: n = 2α

2α^7 + 7 = (2β+1)[p]2[/p] = 4β[p]2[/p]+4β+1

2α^7 = 2(2β[p]2[/p]+2β-3) = 2 X oddno
         hence lhs even powers > rhs even powers imposibble

case 2: n = 2α+1

2α+1^7 - 1 =4β[p]2[/p] - 8

2α( odd) = 4(β[p]2[/p]-2)
α(odd) = 4(β[p]2[/p]/2-1) if  β is odd α=2 , else α=4

so n = 2α+1 has possibly 9,5 has sol
but 5^7 + 7 ends in 2  obviously not square
    9^7 + 7 ends in 6  and is not div by 3 itself so no ending with 6 ruled out

hence all possibilities are ruled out
i have skipped some simple extra explanations in btw as it will make the post too long;)

in your step

a(odd)=4(β[p]2[/p]/2 -1)

when β=even

β=2k

hence we get =
4(2k[p]2[/p]-1)=a(odd)

how come you say that a=4 and no other ?

a can also be 4*(some factor of 2n[p]2[/p]-1)

how come 2n[p]2[/p]-1 is a prime always ?

eg. 2(5)[p]2[/p]-1 = 49=7*7

some one try it ...

did you try induction

i dont think induction will help much

by the way could u prove it by induction..

ya rohan you're absolutely rite
sorry

Please see http://www.mathlinks.ro/viewtopic.php?t=224626

thanks!!

Tried Congruences?

n^7+7 to be a square means n has to be of the form 4n+1 (Obtained by congruences), so effectively if u can prove that 4n+1 does not satisfy this condition, its done.