ANOTHER QUADRATIC

sir,
didnt understandhow will taking diff help.......

then how will u introduce the fact that f(x) = 5 has 4 integrL soln??????

these kind of problems are well known.

f(x) - 5 = (x-n₁) (x-n₂)(x-n₃)(x-n₄) P(x), where P(x) is a polynomial with integer coefficients

If f(n) = 7, then f(n) - 5 = 2 i.e. (n-n₁) (n-n₂)(n-n₃)(n-n₄) P(n) = 2

Now the factors of 2 are ±1, ±2. and remember that only one of ±2 can appear on LHS. Also (n-n₁), (n-n₂), (n-n₃), (n-n₄) are all distinct. By pigeonhole principle, we must two of the roots equal, but that contradicts the problem condition

COOL APPROACH PROPHET!!!!!!!!

hey prophet jus didnt get dat last bit.........

how did u prove dat 4 polynomial = 7 has 0 zero roots.........

in short didnt get it frm here on,

"remember that only one of ±2 can appear on LHS. Also (n-n1), (n-n2), (n-n3), (n-n4) are all distinct."

i read the pigeonhole principle.... got it...... but how did u apply it to our situation????

The numbers 1, -1 and 2 (or -2) are to be assigned to n-n₁, n-n₂, n-n₃ and n-n₄. So some two of them will be equal say n-n₂ and n-n₄. This means n₂ = n₄. But that is forbidden.

why do you say only one out of
+2 and -2 has to be assigned????

i m sorry if this que sounds a lil dumb....

hello mr prophet i want to ask you something many terms that you use in your proofs i havent even seen or heard them before

are they really in the jee level please reply

PROPHET SIR!!!!

I M RAELLY SORRY, I DINNO U WER A TARGETIIT NOBLE!!!!

PL. FORGIVE ME.......