Arithmetic: answer key to the exercises in the final year of high school in PDF.

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Exercise of arithmetic in terminale S.
1-Position d = gcd(a,b)
We have if d divides a and d divides b then d divides b and d divides (a-bq)
Reciprocally: if d divides b and d divides (a-bq) then d divides ( a – bq ) +bq = a
2- it is the previous relationship with a\,=\,5n^3\,-n\,;\,b\,=\,n+2\,;\,q\,=\,5n^2-\,10n\,+19.

Demonstrate a property, exercise of mathematics in terminale S on arithmetic.

Let δ = PGCD(a ;b) and µ = PPCM(a ;b).
Then we have a= δa’ and b = δb’ with a’ and b’ prime between them.
So we have PPCM(a’ ;b’) = a’b’
µ = PPCM(δa’; δb’) = δ×PPCM(a’; b’) = δ×a’×b’
Thus δµ = δ²×a’×b’ = δ×a’× δ×b’ = ab

System of equations and arithmetic.

From the relation PPCM(a;b)×PGCD(a;b) = ab,
we deduce from the second equation of the system :
3×PPCM(a ;b) = PGCD(a ;b)×PPCM(a ;b)
So PGCD(a;b) = 3
Then there are integers a’ and b’ (prime to each other) such that a = 3a’ and b = 3b’.
Carrying over into the first equation, we get :
9a’² – 9b’² = 405
Let (a’ + b’)(a’ – b’) = 45
Now 45 = 3²×5
We can therefore have the following 3 systems:
\,\{\,a'\,+\,b'\,=\,45\\\,a'\,-b'\,=\,1\,. \,\{\,a'\,+\,b'\,=\,15\\\,a'\,-b'\,=\,3\,. \,\{\,a'\,+\,b'\,=\,9\\\,a'\,-b'\,=\,5\,.
a’ = 23 and b’ = 22 a’ = 9 and b’ = 6 a’ = 7 and b’ = 2
Hence (a ;b) = (66 ;69) is not suitable because hence (a ;b) = (42 ;6)
9 and 6 are not prime to each other
The solution pairs of the system are therefore (42 ;6) and (66 ;69).

Prime numbers .

1) (a^2\,+\,2b^2)^2\,=\,a^4\,+\,4a^2b^2\,+\,4b^4
2) a^4+\,4b^4\,=\,(a^2\,+\,2b^2)^2\,-\,4a^2b^2\,=\,(a^2\,+\,2b^2\,+\,2ab)(a^2\,+\,2b^2\,-\,2ab)
a^4+\,4b^4=\,%5B(a\,+\,b)^2\,+\,b^2%5D%5B(a\,-\,b)^2\,+\,b^2%5D
(a\,+\,b)^2\,+\,b^2\geq\,\,4\,et\,(a\,-\,b)^2\,+\,b^2\geq\,\,4. because a and b are greater than or equal to 2.
a^4+4b^4 is the product of two integers greater than 1.
So a^4+4b^4 is not a prime number.

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The exercises in the final year


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Senior year exercises.

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