1in your step
a(odd)=4(β2/2 -1)
when β=even
β=2k
hence we get =
4(2k2-1)=a(odd)
how come you say that a=4 and no other ?
a can also be 4*(some factor of 2n2-1)
how come 2n2-1 is a prime always ?
eg. 2(5)2-1 = 49=7*7
Prove that for any integer n
n7+7 is not a perfect square
i have proved it for all cases except for n of the form - 7k+1
plz help .
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10 Answers
9Rohan heres my sol its quite diff from wat ur expecting
case 1: n = 2α
2α^7 + 7 = (2β+1)2 = 4β2+4β+1
2α^7 = 2(2β2+2β-3) = 2 X oddno
hence lhs even powers > rhs even powers imposibble
case 2: n = 2α+1
2α+1^7 - 1 =4β2 - 8
2α( odd) = 4(β2-2)
α(odd) = 4(β2/2-1) 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;)
1
1i dont think induction will help much
by the way could u prove it by induction..
1n^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.