emptFullMarks":null},{"id":"0b983462-5590-4395-9ed5-1ef9ec2af03e","revisionId":"e31e4a41-9d42-4228-a240-e23d101854f3","questionNumber":"10","questionType":"LA","specification":"","difficulty":2,"difficultyBand":null,"totalMarks":2,"sourceTitle":"Proof","calculatorUse":null,"parts":[{"id":"3f345897-09d5-4178-8d5e-d2407307776f","label":"0","content":"Prove algebraically that the sum of the squares of any 2 odd positive integers is always even.","markscheme":"- Expression for sum of squares: $(2n + 1)^2 + (2m + 1)^2$ M1\n- Expands and simplifies to $4n^2 + 4n + 4m^2 + 4m + 2$ and factors out 2: $2(2n^2 + 2n + 2m^2 + 2m + 1)$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Writes an algebraic expression for the sum of the squares of two odd integers.","code":"M"},{"id":"p2","marks":1,"description":"Simplifies the expression and factors out 2 to show it is even.","code":"A"}],"finalAnswers":[],"generatedAt":"2026-05-07T02:12:13.876Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"algebraic_parity_proof","label":"Algebraic parity proof","steps":[{"type":"text","order":0,"title":"Represent odd numbers","content":"Any odd positive integer can be written as $2k+1$, where $k$ is a non-negative integer. Therefore, let the two odd positive integers be $2n+1$ and $2m+1$, where $n$ and $m$ are non-negative integers. This sets up the proof; the required sum is written in the next step."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S=\\htmlData{anchor=sum}{\\color{@sky}{(2n+1)^2+(2m+1)^2}}$"},{"latex":"$$S=\\htmlData{anchor=expanded}{\\color{@orange}{4n^2+4n+1+4m^2+4m+1}}$"},{"latex":"$$S=4n^2+4n+4m^2+4m+2$"},{"text":"The first line is the required expression for the sum of the squares, so it earns the M1 mark. Expanding and simplifying to the final line is the first part of the A1 work."}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"sum"},{"color":"orange","style":"text-color","anchor":"expanded"}]},"order":1,"title":"Expand the squares","description":"Write the sum of the squares, then expand each bracket and collect the constant terms."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S=\\color{@sky}{\\htmlData{anchor=unfactored}{4n^2+4n+4m^2+4m+2}}$"},{"latex":"$$S=\\color{@green}{\\htmlData{anchor=factored}{2(2n^2+2n+2m^2+2m+1)}}$"},{"text":"The expansion shown in the previous step and the factorisation shown above together earn the A1 mark. The expression in brackets is an integer, so the sum is 2 multiplied by an integer and is therefore even."}],"highlights":[{"color":"sky","style":"text-color","anchor":"unfactored"},{"color":"green","style":"text-color-with-background","anchor":"factored"}]},"order":2,"title":"Factor out 2","description":"Take the common factor of 2 out of every term."}]}],"generatedAt":"2026-07-20T16:55:38.741Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"0e1344db-9d9a-4049-8613-03673b5bbe76","revisionId":"50d234e3-7f14-43fd-a8d0-22dbad01b07e","questionNumber":"11","questionType":"LA","specification":"","difficulty":2,"difficultyBand":null,"totalMarks":3,"sourceTitle":"Proof","calculatorUse":null,"parts":[{"id":"e8cd9e36-6dbd-4c7f-bd0b-747bcb963cce","label":"0","content":"Prove that the sum of the squares of any two consecutive integers is always an odd number.","markscheme":"- Expression for sum of squares: $n^2 + (n + 1)^2$ M1\n- Expands and simplifies: $n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1$ M1\n- Concludes $2(n^2 + n) + 1$ is even + 1, which is odd A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Writes an algebraic expression for the sum of the squares of two consecutive integers.","code":"M"},{"id":"p2","marks":1,"description":"Correctly expands and simplifies the expression to $2n^2 + 2n + 1$.","code":"M"},{"id":"p3","marks":1,"description":"Provides a concluding

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